The word problem in group theory
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The word problem in group theory
The word problem in group theory
Christian Perfect
April 29, 2008
1 Some group theory
Recall that a group G is a set of elements with a multiplicative operation ?
such that for each g1; g2 2 G , g1 ? g2 = g, with g 2 G; G has an identity
element id (or 1) such that g ? id = id ? g = g ; 8g 2 G and every g has
a unique inverse, g 1, where gg 1 = 1. Note that it is not necessarily true
that a group is commutative, i.e. g1 ? g2 = g2 ? g1.
The trivial group is the group consisting of only one element, the identity.
Its group operation is id ? id = id.
From now on we use juxtaposition to denote multiplication, rather than ?,
i.e. x ? y will be instead written xy.
A group can be de ned by a presentation, which is a set of generators X
and a set of relations R on those generators. We write
G = hX ;Ri:
Every element of a group presented this way can be written as a product
of the generators and their inverses. The relations de ne which strings of
generators are equivalent. In a presentation the set R is almost always a
subset of all of the true relations on the group, from which all of the others
can be deduced. The relations like xx 1 = 1 and 1x = x1 = x, which
hold for every group, called here the free relations, are taken as read in
every presentation, so there is no need to include them in R. Here are some
examples of group presentations:
hx ; ;i the in nite cyclic group;
hx ; x5 = 1i the cyclic group of order 5, or Z5;
hx; y ; xy = yxi the abelian group:
1
Christian Perfect
April 29, 2008
1 Some group theory
Recall that a group G is a set of elements with a multiplicative operation ?
such that for each g1; g2 2 G , g1 ? g2 = g, with g 2 G; G has an identity
element id (or 1) such that g ? id = id ? g = g ; 8g 2 G and every g has
a unique inverse, g 1, where gg 1 = 1. Note that it is not necessarily true
that a group is commutative, i.e. g1 ? g2 = g2 ? g1.
The trivial group is the group consisting of only one element, the identity.
Its group operation is id ? id = id.
From now on we use juxtaposition to denote multiplication, rather than ?,
i.e. x ? y will be instead written xy.
A group can be de ned by a presentation, which is a set of generators X
and a set of relations R on those generators. We write
G = hX ;Ri:
Every element of a group presented this way can be written as a product
of the generators and their inverses. The relations de ne which strings of
generators are equivalent. In a presentation the set R is almost always a
subset of all of the true relations on the group, from which all of the others
can be deduced. The relations like xx 1 = 1 and 1x = x1 = x, which
hold for every group, called here the free relations, are taken as read in
every presentation, so there is no need to include them in R. Here are some
examples of group presentations:
hx ; ;i the in nite cyclic group;
hx ; x5 = 1i the cyclic group of order 5, or Z5;
hx; y ; xy = yxi the abelian group:
1
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A word in G is a string x1x2 : : : xn 1xn, with each xi 2 X, with X =
X [X 1, representing the product x1x2 : : : xn 1xn 2 G. Note that a single
element of G can be represented by more than one word, for example xx 1x
and x represent the same element. When the meaning is ambiguous, we
will use w1 = w2 to mean w1 and w2 are the same word, and w1 =G w2 to
mean that w1 and w2 represent the same element in the group G. When you
have the same generator several times in a row, it is useful for clarity to use
superscript notation, eg. xxyyyx 1x 1 can be written instead as x2y3x 2.
If X is a set of letters, we say X? is the set of words constructed from X,
including the empty word.
It is of course possible to write every relation in the form w = 1, where w
is a word. You can then write R as simply the list of left-hand-sides of the
relations, for example:
G = hx; y; a ; xa = ay; a2 = 1i
can be written instead
G = hx; y; a ; xay 1a 1; a2i
Note that a group presentation is not unique - there can be other presenta-
tions that de ne the same group.
A group is said to be nitely generated if the set of generators is nite. It is
said to be nitely presented if the set of relations is also nite.
A group is free if R is the empty set. It is virtually free if it has a free
subgroup of nite index.
A group H is a subgroup of G if H G and h1; h2 2 H ) h1h
1
2 2 H, i.e.
the inverse of every element in H is in H and the product of every pair of
elements of H is in H.
A subgroup N of G is normal if n 2 N; g 2 G ) gng 1 2 N . The normal
closure fngG of an element n in G is the set fg 1ngj8g 2 Gg.
The left cosets of a subgroup H in G are the sets gH = fgh ; h 2 Hg, g 2 G.
Clearly, by the de nition, every element of G belongs to a unique left coset.
Similarly, the right cosets of H in G are the sets Hg = fhg ; h 2 Hg, g 2 G.
The index of H in G, written [G : H], is the number of distinct left (or right)
cosets.
Given a subgroup H of G, the quotient group G=H of G over H is de ned
to be the set of all left cosets of H in G, with the group operation de ned
as (aH)(bH) = (ab)H, for each aH; bH 2 G=H.
2
X [X 1, representing the product x1x2 : : : xn 1xn 2 G. Note that a single
element of G can be represented by more than one word, for example xx 1x
and x represent the same element. When the meaning is ambiguous, we
will use w1 = w2 to mean w1 and w2 are the same word, and w1 =G w2 to
mean that w1 and w2 represent the same element in the group G. When you
have the same generator several times in a row, it is useful for clarity to use
superscript notation, eg. xxyyyx 1x 1 can be written instead as x2y3x 2.
If X is a set of letters, we say X? is the set of words constructed from X,
including the empty word.
It is of course possible to write every relation in the form w = 1, where w
is a word. You can then write R as simply the list of left-hand-sides of the
relations, for example:
G = hx; y; a ; xa = ay; a2 = 1i
can be written instead
G = hx; y; a ; xay 1a 1; a2i
Note that a group presentation is not unique - there can be other presenta-
tions that de ne the same group.
A group is said to be nitely generated if the set of generators is nite. It is
said to be nitely presented if the set of relations is also nite.
A group is free if R is the empty set. It is virtually free if it has a free
subgroup of nite index.
A group H is a subgroup of G if H G and h1; h2 2 H ) h1h
1
2 2 H, i.e.
the inverse of every element in H is in H and the product of every pair of
elements of H is in H.
A subgroup N of G is normal if n 2 N; g 2 G ) gng 1 2 N . The normal
closure fngG of an element n in G is the set fg 1ngj8g 2 Gg.
The left cosets of a subgroup H in G are the sets gH = fgh ; h 2 Hg, g 2 G.
Clearly, by the de nition, every element of G belongs to a unique left coset.
Similarly, the right cosets of H in G are the sets Hg = fhg ; h 2 Hg, g 2 G.
The index of H in G, written [G : H], is the number of distinct left (or right)
cosets.
Given a subgroup H of G, the quotient group G=H of G over H is de ned
to be the set of all left cosets of H in G, with the group operation de ned
as (aH)(bH) = (ab)H, for each aH; bH 2 G=H.
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If there are t distinct left (or right) cosets in G=H, the coset representatives
for H in G are a selection of elements b1; b2; : : : ; bt such that each bi belongs
to a di erent coset.
Lemma 1.1 If we are given a group G, a nite index subgroup H of G with
generating set Y and a word w, then as w is read, one letter at a time, it
can be rephrased as a word in the form wyb, where wy is a word in Y ? and
b 2 B where B is the set of the coset representatives of H in G.
Proof: Given two di erent generating sets for G, we can rewrite the gener-
ators of one as words from the other, so we are free to choose the generating
set we will use. Let X = Y [B be the generating set for G. Start with the
empty word, , and add to it the rst letter of w. No matter what letter is
read, this word is in the desired form.
Now suppose we have already read a portion of w and it has been rephrased
as a word wyb, for some wy 2 Y ?, b 2 B.The next letter we read will either
be a member of Y or B.
First suppose it is some y 2 Y . Then we have the word wyby. by belongs
to some right coset Hb, so is equivalent to some w0yb
0. So the word we have
read is equivalent to wyw0yb
0, which is in the desired form.
Suppose instead that we read some b0 2 B. Then clearly the resulting word,
wbb0, is equivalent to some wb00, which is in the desired form.
If we have an element g 2 G, the conjugate of g by x 2 G is de ned to
be xgx 1. Conjugation is a homomorphism, that is, the product of the
conjugations of two elements is equal to the conjugation of the product of the
elements. This is easy to show: assume we have g; h 2 G and conjugate by
x: xgx 1xhx 1 = xghx 1. An important fact to note is that the conjugate
of the identity element is the identity element itself, and the conjugate of
any nonidentity element must be a nonidentity element.
2 The word problem
Recall that, given a nitely generated group G = hX ;Ri, the set of words
is the set of strings w of generators and their inverses x 2 X. We denote
by the empty word.
For example, for the group G = hx; y; a ; xa = ay; a2 = 1i, some words are
xyx 1; x2y; a4;
3
for H in G are a selection of elements b1; b2; : : : ; bt such that each bi belongs
to a di erent coset.
Lemma 1.1 If we are given a group G, a nite index subgroup H of G with
generating set Y and a word w, then as w is read, one letter at a time, it
can be rephrased as a word in the form wyb, where wy is a word in Y ? and
b 2 B where B is the set of the coset representatives of H in G.
Proof: Given two di erent generating sets for G, we can rewrite the gener-
ators of one as words from the other, so we are free to choose the generating
set we will use. Let X = Y [B be the generating set for G. Start with the
empty word, , and add to it the rst letter of w. No matter what letter is
read, this word is in the desired form.
Now suppose we have already read a portion of w and it has been rephrased
as a word wyb, for some wy 2 Y ?, b 2 B.The next letter we read will either
be a member of Y or B.
First suppose it is some y 2 Y . Then we have the word wyby. by belongs
to some right coset Hb, so is equivalent to some w0yb
0. So the word we have
read is equivalent to wyw0yb
0, which is in the desired form.
Suppose instead that we read some b0 2 B. Then clearly the resulting word,
wbb0, is equivalent to some wb00, which is in the desired form.
If we have an element g 2 G, the conjugate of g by x 2 G is de ned to
be xgx 1. Conjugation is a homomorphism, that is, the product of the
conjugations of two elements is equal to the conjugation of the product of the
elements. This is easy to show: assume we have g; h 2 G and conjugate by
x: xgx 1xhx 1 = xghx 1. An important fact to note is that the conjugate
of the identity element is the identity element itself, and the conjugate of
any nonidentity element must be a nonidentity element.
2 The word problem
Recall that, given a nitely generated group G = hX ;Ri, the set of words
is the set of strings w of generators and their inverses x 2 X. We denote
by the empty word.
For example, for the group G = hx; y; a ; xa = ay; a2 = 1i, some words are
xyx 1; x2y; a4;
3
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We might want to know which words are equivalent to the identity element
of G. This is called the word problem on G. 'Word problem' also refers to
the set of words equivalent to the identity, written WP (G). In the group G
given before, some words in WP (G) are:
aa 1; x2ay 1ax 1; xa2x 1
It would be interesting to be able to say for a given word w whether it is in
WP (G) or not, or to be able to construct all of WP (G) for a given group.
Clearly the former problem is a computation problem, so we would expect
to be able to construct a machine for a given group that would take words
as its input and either accept them if they are equivalent to the identity
element or reject them if they are not.
3 Automaton Theory
3.1 Finite State Automata
A deterministic nite state automaton [11] or FSA is a machine which takes
as input a string w of letters from an alphabet . It begins in a starting
state i and at each state s reads and deletes the leftmost character x of w
and changes to the state (s; x). If the machine is in an accepting state when
it reaches the last letter of w it accepts w, otherwise it rejects w. An FSA
can be thought of as a directed graph with the nodes representing states
and the edges labelled by letters. You can then say a word is accepted if it
labels a path from the start state to an accepting state.
A non-deterministic nite state automaton is like a deterministic FSA but
instead of just one starting state there is a set i of starting states, and from
each state (s; x) is a set of one or more destination states. So then a word
is accepted if it labels one or more paths from some starting state to any
accepting state.
Formally, a nite state automaton is a 5-tuple M = (Q;; ; i; F ); where Q
is a set of states, is an alphabet, is a map : Q ! Q, i is a set of
initial states and F is a set of accepting states.
Lemma 3.1 The word problem of a nite group can be solved on a nite
state automaton. [12]
Proof: For a given nite group G generated by the (necessarily nite) set
X, construct an FSA with Q = G, = X, i = 1, F = 1 and (g; x) = gx.
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of G. This is called the word problem on G. 'Word problem' also refers to
the set of words equivalent to the identity, written WP (G). In the group G
given before, some words in WP (G) are:
aa 1; x2ay 1ax 1; xa2x 1
It would be interesting to be able to say for a given word w whether it is in
WP (G) or not, or to be able to construct all of WP (G) for a given group.
Clearly the former problem is a computation problem, so we would expect
to be able to construct a machine for a given group that would take words
as its input and either accept them if they are equivalent to the identity
element or reject them if they are not.
3 Automaton Theory
3.1 Finite State Automata
A deterministic nite state automaton [11] or FSA is a machine which takes
as input a string w of letters from an alphabet . It begins in a starting
state i and at each state s reads and deletes the leftmost character x of w
and changes to the state (s; x). If the machine is in an accepting state when
it reaches the last letter of w it accepts w, otherwise it rejects w. An FSA
can be thought of as a directed graph with the nodes representing states
and the edges labelled by letters. You can then say a word is accepted if it
labels a path from the start state to an accepting state.
A non-deterministic nite state automaton is like a deterministic FSA but
instead of just one starting state there is a set i of starting states, and from
each state (s; x) is a set of one or more destination states. So then a word
is accepted if it labels one or more paths from some starting state to any
accepting state.
Formally, a nite state automaton is a 5-tuple M = (Q;; ; i; F ); where Q
is a set of states, is an alphabet, is a map : Q ! Q, i is a set of
initial states and F is a set of accepting states.
Lemma 3.1 The word problem of a nite group can be solved on a nite
state automaton. [12]
Proof: For a given nite group G generated by the (necessarily nite) set
X, construct an FSA with Q = G, = X, i = 1, F = 1 and (g; x) = gx.
4
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x
1
Figure 1: A nite state automaton accepting the word problem of the cyclic
group Z6 = hx ; x6 = 1i
. The solid dot represents the accepting state.
3.2 Push-down Automata
A push-down automaton [11] is a FSA with an additional stack which is read
from and added to during each move. The stack is a list of letters, which
the machine can only read the last letter from or add a letter (or string of
letters) to the end of. A PDA can then at each stage make a move like a
FSA, reading a letter from the input and depending on that, the current
state and the last letter of the stack, move to a di erent state and/or add
a new string to the stack. It can also make moves where nothing is read
from the input and what happens depends only on the current state and the
stack. In this case we say the empty string was read from the input, and
an -move is made. The PDA accepts the input if it is in an accepting state
with the stack empty after it has read the last letter of the input. Otherwise
it rejects the input.
So a PDA is a 6-tuple M = (Q;; ; ; i; F ) where Q, , i and F are as
before, but is the alphabet of letters that can be put on the stack, and
is rede ned as : Q [ fg ! Q ?.
Lemma 3.2 The word problem of a free group can be solved on a pushdown
automaton.
Proof: Given a free groupG = hX ; ;i, every word is of the form xa11 x
a2
2 : : : x
ann
with xi 2 X [X 1 for i = 1 : : : n. It is easy to see that a word w represents
the identity element of G if and only if w can be reduced to the empty string
by successive deletions of subwords of the form xx 1, x 2 X [X 1. Create
a PDA with only one state, which is both the initial and accepting state.
At each move, read a letter x from the input. If x 1 is on the end of the
stack, delete it, otherwise add x to the stack.
5
1
Figure 1: A nite state automaton accepting the word problem of the cyclic
group Z6 = hx ; x6 = 1i
. The solid dot represents the accepting state.
3.2 Push-down Automata
A push-down automaton [11] is a FSA with an additional stack which is read
from and added to during each move. The stack is a list of letters, which
the machine can only read the last letter from or add a letter (or string of
letters) to the end of. A PDA can then at each stage make a move like a
FSA, reading a letter from the input and depending on that, the current
state and the last letter of the stack, move to a di erent state and/or add
a new string to the stack. It can also make moves where nothing is read
from the input and what happens depends only on the current state and the
stack. In this case we say the empty string was read from the input, and
an -move is made. The PDA accepts the input if it is in an accepting state
with the stack empty after it has read the last letter of the input. Otherwise
it rejects the input.
So a PDA is a 6-tuple M = (Q;; ; ; i; F ) where Q, , i and F are as
before, but is the alphabet of letters that can be put on the stack, and
is rede ned as : Q [ fg ! Q ?.
Lemma 3.2 The word problem of a free group can be solved on a pushdown
automaton.
Proof: Given a free groupG = hX ; ;i, every word is of the form xa11 x
a2
2 : : : x
ann
with xi 2 X [X 1 for i = 1 : : : n. It is easy to see that a word w represents
the identity element of G if and only if w can be reduced to the empty string
by successive deletions of subwords of the form xx 1, x 2 X [X 1. Create
a PDA with only one state, which is both the initial and accepting state.
At each move, read a letter x from the input. If x 1 is on the end of the
stack, delete it, otherwise add x to the stack.
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Lemma 3.3 The word problem of a nitely generated virtually free group
can be solved on a pushdown automaton. [12, 13]
1
a
a
x
y
x
y
+x
+y
+y
+x
Figure 2: A pushdown automaton accepting the word problem of the vir-
tually free group G = hx; y; a ; xa = ay; a2 = 1i. Dotted lines labelled by a
letter z1 and ending in +z2 indicate a move adding z2 to the stack after z1
is read from the input.
Proof: Let G be a virtually free group with a free subgroup H of nite
index. It is well-known that any subgroup of G of nite index contains a
subgroup N which is normal in G and itself has nite index. Furthermore, a
theorem proved by Schreier says that any subgroup having nite index in a
nitely generated group is itself nitely generated, so N is nitely generated
by a set Y (which we can express as a set of words over X).
Let B = G=N , the quotient group of G over N . Let y1; : : : ; yn be the free
generators of N , and b0; : : : ; bt be the coset representatives for N in G, with
b0 representing the identity element of B. To solve the word problem for a
word w, create a PDA with a state qi for each bi, to keep track of the image
of w in B, and where q0 is the initial and accepting state.
So at any one stage in the running of our PDA, the word we have read
is equivalent to one of the form wyb, where wy 2 Y ? (w freely reduced,
i.e. shortened as much as it can be using only the free relations) and b 2
fb0; b1; : : : ; btg, meaning at that stage we have wy on the stack and are in
the state labelled by b. Now suppose we read a letter x. The word wybx is
equal to a word wyw0yb
0, w0y 2 Y
?, b0 2 fb0; b1; : : : ; btg. Using only the stack
we can nd the free reduction of wyw0y and end up with that on stack. We
also move to the state representing b0. If, once we have read all of the input
6
can be solved on a pushdown automaton. [12, 13]
1
a
a
x
y
x
y
+x
+y
+y
+x
Figure 2: A pushdown automaton accepting the word problem of the vir-
tually free group G = hx; y; a ; xa = ay; a2 = 1i. Dotted lines labelled by a
letter z1 and ending in +z2 indicate a move adding z2 to the stack after z1
is read from the input.
Proof: Let G be a virtually free group with a free subgroup H of nite
index. It is well-known that any subgroup of G of nite index contains a
subgroup N which is normal in G and itself has nite index. Furthermore, a
theorem proved by Schreier says that any subgroup having nite index in a
nitely generated group is itself nitely generated, so N is nitely generated
by a set Y (which we can express as a set of words over X).
Let B = G=N , the quotient group of G over N . Let y1; : : : ; yn be the free
generators of N , and b0; : : : ; bt be the coset representatives for N in G, with
b0 representing the identity element of B. To solve the word problem for a
word w, create a PDA with a state qi for each bi, to keep track of the image
of w in B, and where q0 is the initial and accepting state.
So at any one stage in the running of our PDA, the word we have read
is equivalent to one of the form wyb, where wy 2 Y ? (w freely reduced,
i.e. shortened as much as it can be using only the free relations) and b 2
fb0; b1; : : : ; btg, meaning at that stage we have wy on the stack and are in
the state labelled by b. Now suppose we read a letter x. The word wybx is
equal to a word wyw0yb
0, w0y 2 Y
?, b0 2 fb0; b1; : : : ; btg. Using only the stack
we can nd the free reduction of wyw0y and end up with that on stack. We
also move to the state representing b0. If, once we have read all of the input
6
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and reduced the stack, we are in the state q0 with an empty stack, we can
say the input is equivalent to the identity and it is accepted.
In fact, the converse of this lemma is also true; if you can solve the word
problem of a group on a pushdown automaton then it is necessarily virtually
free. The proof of this is considerably longer than the previous lemma, and
requires knowledge of context-free languages.
4 Languages and grammars
Since we're talking a lot about words, it would be useful to have a more
formal de nition of what they are and how they can be made. Following
Hopcroft and Ullman [11], we formally de ne a language to be a set of words
whch are made of letters from a certain alphabet and which obey a set of
syntax rules specifying which words are valid members of the language.
Rather than using the syntax to see which words belong to a language, we
could de ne a grammar which describes how to construct all of a language's
words. Formally, a grammar is a set of production rules describing how
strings of symbols can be transformed. To generate a word, start with a start
symbol and then successively apply production rules to change the string.
The symbols in a grammar are either terminal or nonterminal. Terminal
symbols can not be transformed into any other strings.
Formally, a grammar consists of a set N of nonterminal symbols (also called
variables) including a single start symbol S, a set of terminal symbols
disjoint from N , and a set of production rules of the form ( [N)?N( [
N)? ! ([N)?, that is, the left-hand side of every production rule contains
at least a nonterminal symbol, but the right hand side is allowed to be the
empty string (represented by ).
A context-free grammar is one in which the left-hand side of every production
rule is a single nonterminal symbol. It is called context-free because every
rule can be applied wherever its left-hand side symbol appears, regardless of
the rest of the string. A language is said to be context-free if it is generated
by a context-free grammar.
A context-free grammar is said to be in Greibach normal form if all of
the production rules are of the form A ! X or S ! , where A is a
nonterminal symbol, is a terminal symbol and X is a (possibly empty)
string of nonterminal symbols.
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say the input is equivalent to the identity and it is accepted.
In fact, the converse of this lemma is also true; if you can solve the word
problem of a group on a pushdown automaton then it is necessarily virtually
free. The proof of this is considerably longer than the previous lemma, and
requires knowledge of context-free languages.
4 Languages and grammars
Since we're talking a lot about words, it would be useful to have a more
formal de nition of what they are and how they can be made. Following
Hopcroft and Ullman [11], we formally de ne a language to be a set of words
whch are made of letters from a certain alphabet and which obey a set of
syntax rules specifying which words are valid members of the language.
Rather than using the syntax to see which words belong to a language, we
could de ne a grammar which describes how to construct all of a language's
words. Formally, a grammar is a set of production rules describing how
strings of symbols can be transformed. To generate a word, start with a start
symbol and then successively apply production rules to change the string.
The symbols in a grammar are either terminal or nonterminal. Terminal
symbols can not be transformed into any other strings.
Formally, a grammar consists of a set N of nonterminal symbols (also called
variables) including a single start symbol S, a set of terminal symbols
disjoint from N , and a set of production rules of the form ( [N)?N( [
N)? ! ([N)?, that is, the left-hand side of every production rule contains
at least a nonterminal symbol, but the right hand side is allowed to be the
empty string (represented by ).
A context-free grammar is one in which the left-hand side of every production
rule is a single nonterminal symbol. It is called context-free because every
rule can be applied wherever its left-hand side symbol appears, regardless of
the rest of the string. A language is said to be context-free if it is generated
by a context-free grammar.
A context-free grammar is said to be in Greibach normal form if all of
the production rules are of the form A ! X or S ! , where A is a
nonterminal symbol, is a terminal symbol and X is a (possibly empty)
string of nonterminal symbols.
7
Page 8
A context-free grammar is said to be in Chomsky normal form if all of the
production rules are of the form A! BC, A! or S ! , where A,B and
C are nonterminal symbols and is a terminal symbol.
Sometimes these kinds of grammars are de ned not to include rules of the
form A! , so that applications of production rules can only make a string
longer or the same length, making some proofs a little easier. However, we
need to be in our languages, so we will not follow this convention.
If A
is a string in a language produced by a context-free grammar C, and
A! is a production rule in C, we say there is a derivation A
)
.
If there is a series of derivations ) 1; 1 ) 2; : : : ; m 1 ) m; m ) ,
we write
?
=) .
A variable A is said to be useful if there is a derivation S
?
=) A
?
=) w
of a string of terminals which uses A. If A is not useful, it is useless. The
language L(A) generated by a variable A is de ned to be the set of words
which are derivable from A, that is, L(A) = fwjw 2 ?andA
?
=) wg. L(A)
is non-empty if A is useful.
A reduced grammar is one which contains only useful variables. Every
context-free language can be generated by a grammar in Greibach normal
form or Chomsky normal form.
Lemma 4.1 The language generated by a context-free grammar C can also
be generated exactly by a context-free grammar C 0 in Greibach normal form.
Proof: Consider a derivation A ! w, with w = w1 : : : wn 2 ( [ N)?.
First of all, if w1 is a nonterminal symbol, rewrite w as w. Suppose wn
is a nonterminal symbol. Then we can write w = w1 : : : wmN1 : : : Nl, with
Ni 2 N for i = 1 : : : l and wm 2 . Rewrite the derivation as
A! BC
B ! w1 : : : wm 1
C ! wmN1 : : : Nl:
Now suppose instead that wn is a terminal symbol . Rewrite the derivation
as
A! BC
B ! w1 : : : wn 1
C ! :
In both these cases, the derivations from A and C are put in Greibach
normal form, and we can repeat the procedure on the shorter word derived
8
production rules are of the form A! BC, A! or S ! , where A,B and
C are nonterminal symbols and is a terminal symbol.
Sometimes these kinds of grammars are de ned not to include rules of the
form A! , so that applications of production rules can only make a string
longer or the same length, making some proofs a little easier. However, we
need to be in our languages, so we will not follow this convention.
If A
is a string in a language produced by a context-free grammar C, and
A! is a production rule in C, we say there is a derivation A
)
.
If there is a series of derivations ) 1; 1 ) 2; : : : ; m 1 ) m; m ) ,
we write
?
=) .
A variable A is said to be useful if there is a derivation S
?
=) A
?
=) w
of a string of terminals which uses A. If A is not useful, it is useless. The
language L(A) generated by a variable A is de ned to be the set of words
which are derivable from A, that is, L(A) = fwjw 2 ?andA
?
=) wg. L(A)
is non-empty if A is useful.
A reduced grammar is one which contains only useful variables. Every
context-free language can be generated by a grammar in Greibach normal
form or Chomsky normal form.
Lemma 4.1 The language generated by a context-free grammar C can also
be generated exactly by a context-free grammar C 0 in Greibach normal form.
Proof: Consider a derivation A ! w, with w = w1 : : : wn 2 ( [ N)?.
First of all, if w1 is a nonterminal symbol, rewrite w as w. Suppose wn
is a nonterminal symbol. Then we can write w = w1 : : : wmN1 : : : Nl, with
Ni 2 N for i = 1 : : : l and wm 2 . Rewrite the derivation as
A! BC
B ! w1 : : : wm 1
C ! wmN1 : : : Nl:
Now suppose instead that wn is a terminal symbol . Rewrite the derivation
as
A! BC
B ! w1 : : : wn 1
C ! :
In both these cases, the derivations from A and C are put in Greibach
normal form, and we can repeat the procedure on the shorter word derived
8
Page 9
from B to produce a set of derivations in Greibach normal form equivalent
to the one given.
Lemma 4.2 The language generated by a context-free grammar C can also
be generated exactly by a context-free grammar C 0 in Chomsky normal form.
Proof: Consider a derivation A ! w, with w = w1 : : : wn 2 ( [ N)?.
Suppose that wn is a terminal symbol. Rewrite the derivation as
A! BC
B ! w1 : : : wn 1
C ! wn:
Now suppose instead that w1 is a terminal symbol. Rewrite the derivation
as
A! CB
B ! w2 : : : wn
C ! w1:
Suppose that w = N1 : : : Nn, with Ni 2 N for i = 1 : : : n. Rewrite the
derivation as
A! N1A2
A2 ! N2A3
: : :
Ai ! NiAi+1
An = Nn:
Finally, suppose that w = Xw0Y , where X and Y are (possibly empty)
strings of nonterminal symbols, and w0 2 ( [N)?. Rewrite the derivation
as
A! BC
B ! X
C ! DE
D ! w0
E ! Y:
With B ! X and E ! Y rewritten as in the previous case.
9
to the one given.
Lemma 4.2 The language generated by a context-free grammar C can also
be generated exactly by a context-free grammar C 0 in Chomsky normal form.
Proof: Consider a derivation A ! w, with w = w1 : : : wn 2 ( [ N)?.
Suppose that wn is a terminal symbol. Rewrite the derivation as
A! BC
B ! w1 : : : wn 1
C ! wn:
Now suppose instead that w1 is a terminal symbol. Rewrite the derivation
as
A! CB
B ! w2 : : : wn
C ! w1:
Suppose that w = N1 : : : Nn, with Ni 2 N for i = 1 : : : n. Rewrite the
derivation as
A! N1A2
A2 ! N2A3
: : :
Ai ! NiAi+1
An = Nn:
Finally, suppose that w = Xw0Y , where X and Y are (possibly empty)
strings of nonterminal symbols, and w0 2 ( [N)?. Rewrite the derivation
as
A! BC
B ! X
C ! DE
D ! w0
E ! Y:
With B ! X and E ! Y rewritten as in the previous case.
9
Page 10
By repeating application of these four procedures, we can produce a set of
derivations in Chomsky normal form equivalent to the one given.
A regular grammar is one in which all the left-hand sides of production
rules are single nonterminal symbols, like before, but with the additional
constraint on the right-hand sides that they must be either a single terminal
symbol, a single terminal symbol followed by a nonterminal symbol, or the
empty string. We say a language is regular if it is generated by a regular
grammar.
A language L is regular if and only if it is accepted by some FSA. It has
been shown that a language L is context-free if and only if L is accepted by
some PDA. For conciseness, we call the language accepted by an automaton
its language. So the language of a FSA is regular, and the language of a
PDA is context-free.
We can consider the word problem of a group (in the sense meaning the set
of words equivalent to the group's identity) to be a language. Now if we
know the grammar generating the language has certain properties, we can
show the group has certain properties, and vice versa.
Lemma 4.3 A group has regular word problem if and only if it is nite.
[12, 1]
Proof: We have already proved in Lemma 3.1 that the word problem of
a nite group can be solved on a nite state automaton. The language
of accepted words of a FSA is regular, so a nite group has regular word
problem. We now need to prove that a group with regular word problem is
nite.
Suppose we have a nitely generated in nite group G = hX ;Ri. For any
integer k, there must be in nitely many elements of G which can not be
represented by a word shorter than k, since if there were nitely many,
and since X is nite, there would be nitely many elements of G. Let M
be a FSA with input alphabet X and n states. Let w be a word with
length greater than n and such that no subword of w represents the identity
element, that is, w can not be made any shorter. Since there are more letters
in w than there are states in M , there must be two initial segments u and
uv of w such that M is in the same state after reading either of them. Now
uu 1 = 1 in G, but uvu 1 6= 1 in G because uvu 1 is a conjugate of v, which
is not equivalent to the identity by the de nition of w. Since M must either
accept or reject both uu 1 and uvu 1 together, the set of words accepted
by M is not exactly equal to the word problem of G.
Clearly, since for a single group there can be several equivalent but distinct
10
derivations in Chomsky normal form equivalent to the one given.
A regular grammar is one in which all the left-hand sides of production
rules are single nonterminal symbols, like before, but with the additional
constraint on the right-hand sides that they must be either a single terminal
symbol, a single terminal symbol followed by a nonterminal symbol, or the
empty string. We say a language is regular if it is generated by a regular
grammar.
A language L is regular if and only if it is accepted by some FSA. It has
been shown that a language L is context-free if and only if L is accepted by
some PDA. For conciseness, we call the language accepted by an automaton
its language. So the language of a FSA is regular, and the language of a
PDA is context-free.
We can consider the word problem of a group (in the sense meaning the set
of words equivalent to the group's identity) to be a language. Now if we
know the grammar generating the language has certain properties, we can
show the group has certain properties, and vice versa.
Lemma 4.3 A group has regular word problem if and only if it is nite.
[12, 1]
Proof: We have already proved in Lemma 3.1 that the word problem of
a nite group can be solved on a nite state automaton. The language
of accepted words of a FSA is regular, so a nite group has regular word
problem. We now need to prove that a group with regular word problem is
nite.
Suppose we have a nitely generated in nite group G = hX ;Ri. For any
integer k, there must be in nitely many elements of G which can not be
represented by a word shorter than k, since if there were nitely many,
and since X is nite, there would be nitely many elements of G. Let M
be a FSA with input alphabet X and n states. Let w be a word with
length greater than n and such that no subword of w represents the identity
element, that is, w can not be made any shorter. Since there are more letters
in w than there are states in M , there must be two initial segments u and
uv of w such that M is in the same state after reading either of them. Now
uu 1 = 1 in G, but uvu 1 6= 1 in G because uvu 1 is a conjugate of v, which
is not equivalent to the identity by the de nition of w. Since M must either
accept or reject both uu 1 and uvu 1 together, the set of words accepted
by M is not exactly equal to the word problem of G.
Clearly, since for a single group there can be several equivalent but distinct
10
Page 11
presentations, the language representing the group's word problem depends
on which presentation is used. Thankfully, the following lemma means we
don't need to worry too much about which particular presentation is used.
Lemma 4.4 Suppose we have a group G with a nitely generated subgroup
H. If WP (G) is a context-free language in one nitely generated presen-
tation of G, then WP (H) is context-free. Hence, WP (G) is a context-free
language in every nitely generated presentation of G.[12]
Proof: Let hX ;Ri be a nitely generated presentation of G with context-
free word problem. Let M be a PDA accepting the word problem of the
latter presentation of G. Let hY ; Si, with Y = y1; : : : ; yn, be a nitely
generated presentation of a subgroup H of G. Then make an embedding
(yi) = ui, i = 1; : : : ; n, where each ui is a word on X. Clearly a word w
on Y only represents the identity element if (w) represents the identity
element in hX ;Ri. So we just need to create a PDA M 0 for WP (H) which
on reading yi simulates what M would do on reading ui. In the case that
H = G with a di erent presentation, WP (H) is context-free, soG is context-
free no matter how it is presented.
There is a more general result
Knowing that a group is context-free tells us something about the structure
of the words in its word problem. We can represent the construction of words
on a graph speci c to the group's presentation, called its Cayley graph, and
consider the graph's geometric properties in order to know more about the
group.
5 The Cayley graph of a presentation group, and
graph triangulations
If G = hX ;Ri is a nitely generated group, the Cayley graph (G) of the
presentation is de ned like so:
For each element of G there is a vertex in the graph. The origin of the
graph is the vertex representing the identity element. For each element
g 2 G and each generator x 2 X, there is a directed edge linking the
vertex representing g to that representing gx, labelled by x.
From the de nition of (G), a word = x1x2 : : : xn is equivalent to the
identity if and only if it labels a closed path in the Cayley graph. So if
11
on which presentation is used. Thankfully, the following lemma means we
don't need to worry too much about which particular presentation is used.
Lemma 4.4 Suppose we have a group G with a nitely generated subgroup
H. If WP (G) is a context-free language in one nitely generated presen-
tation of G, then WP (H) is context-free. Hence, WP (G) is a context-free
language in every nitely generated presentation of G.[12]
Proof: Let hX ;Ri be a nitely generated presentation of G with context-
free word problem. Let M be a PDA accepting the word problem of the
latter presentation of G. Let hY ; Si, with Y = y1; : : : ; yn, be a nitely
generated presentation of a subgroup H of G. Then make an embedding
(yi) = ui, i = 1; : : : ; n, where each ui is a word on X. Clearly a word w
on Y only represents the identity element if (w) represents the identity
element in hX ;Ri. So we just need to create a PDA M 0 for WP (H) which
on reading yi simulates what M would do on reading ui. In the case that
H = G with a di erent presentation, WP (H) is context-free, soG is context-
free no matter how it is presented.
There is a more general result
Knowing that a group is context-free tells us something about the structure
of the words in its word problem. We can represent the construction of words
on a graph speci c to the group's presentation, called its Cayley graph, and
consider the graph's geometric properties in order to know more about the
group.
5 The Cayley graph of a presentation group, and
graph triangulations
If G = hX ;Ri is a nitely generated group, the Cayley graph (G) of the
presentation is de ned like so:
For each element of G there is a vertex in the graph. The origin of the
graph is the vertex representing the identity element. For each element
g 2 G and each generator x 2 X, there is a directed edge linking the
vertex representing g to that representing gx, labelled by x.
From the de nition of (G), a word = x1x2 : : : xn is equivalent to the
identity if and only if it labels a closed path in the Cayley graph. So if
11
Page 12
we can nd the closed paths in the Cayley graph, we can nd the words in
WP (G).
a
a
x y
a
a
a−1
x−1
y
1
Figure 3: The Cayley graph of the group G = hx; y; a ; xa = ay; a2 = 1i
A polygon P is a set of vertices fp0; : : : ; png with a boundary which is a
simple closed curve, that is a series of edges linking p0 to p1, p1 to p2 and so
on, and also pn back to p0. A triangulation of a polygon P is an addition of
edges and possibly vertices so that every polygon contained within P can be
broken down into triangles. A diagonal triangulation of P is a triangulation
which uses only the vertices originally in P . For ease of mind, polygons of
just one or two points are also considered to be diagonally triangulated.
Let G = hX ;Ri be a nitely generated group. Let be a closed path in
(G), labelled by w = y1y2 : : : yn. Write the letters of w clockwise around
the boundary of a regular n-gon P , so that the edges of P are labelled by
the letters of w. A K-triangulation of is a diagonal triangulation of P
with a label from the free group F = hXi assigned to each new edge such
that reading around the boundary of each triangle gives a true relation in
G, and if u is the label on an edge of the triangulation, then juj K.
Lemma 5.1 Let C be a reduced context-free grammar which generates WP (G)
for G = hX ;Ri. If u; v 2 L(A) for some variable A of C, then u and v
represent the same element of G. [12]
Proof: Since C is a reduced grammar, A is useful, so there is a derivation
S
?
=) A
?
=) w1w2w3, where w1w2w3 2WP (G), i.e. w1w2w3 = 1 and w2 is
the part derived from A, i.e. A
?
=) w2. If we replace the derivation of w2 by
the derivations of u and v, we can get S
?
=) w1uw3 and S
?
=) w1vw3, which
12
WP (G).
a
a
x y
a
a
a−1
x−1
y
1
Figure 3: The Cayley graph of the group G = hx; y; a ; xa = ay; a2 = 1i
A polygon P is a set of vertices fp0; : : : ; png with a boundary which is a
simple closed curve, that is a series of edges linking p0 to p1, p1 to p2 and so
on, and also pn back to p0. A triangulation of a polygon P is an addition of
edges and possibly vertices so that every polygon contained within P can be
broken down into triangles. A diagonal triangulation of P is a triangulation
which uses only the vertices originally in P . For ease of mind, polygons of
just one or two points are also considered to be diagonally triangulated.
Let G = hX ;Ri be a nitely generated group. Let be a closed path in
(G), labelled by w = y1y2 : : : yn. Write the letters of w clockwise around
the boundary of a regular n-gon P , so that the edges of P are labelled by
the letters of w. A K-triangulation of is a diagonal triangulation of P
with a label from the free group F = hXi assigned to each new edge such
that reading around the boundary of each triangle gives a true relation in
G, and if u is the label on an edge of the triangulation, then juj K.
Lemma 5.1 Let C be a reduced context-free grammar which generates WP (G)
for G = hX ;Ri. If u; v 2 L(A) for some variable A of C, then u and v
represent the same element of G. [12]
Proof: Since C is a reduced grammar, A is useful, so there is a derivation
S
?
=) A
?
=) w1w2w3, where w1w2w3 2WP (G), i.e. w1w2w3 = 1 and w2 is
the part derived from A, i.e. A
?
=) w2. If we replace the derivation of w2 by
the derivations of u and v, we can get S
?
=) w1uw3 and S
?
=) w1vw3, which
12
Page 13
implies w1uw3 = 1 = w1vw3 in G, since C generates the word problem of
G. Now since G is a group, we have u = v.
From this result, a theorem about the closed paths in the Cayley graph can
be proved.
Theorem 5.2 If a nitely generated group G = hX ;Ri is context-free,
there exists a constant K such that every closed path in the Cayley graph
(G) can be K-triangulated. [12]
Proof: Let G be context-free, and let C be a reduced grammar in Chomsky
normal form which generates WP (G).
Let be a closed path in (G), labelled by w = y1; : : : ; yn, enclosing an
n-gon P . We will create a diagonal triangulation of P by adding edges
labelled by words (not letters) in X joining vertices of P so that the label
of the boundaries of any resulting polygons are true relations in G. The
result is trivial if n = 3, so assume n 4. If A is a variable of C, let uA
denote the shortest word derivable from A. Consider any derivation S
?
=) w.
Since C is in Chomsky normal form, this derivation must be of the form
S ) AB
?
=) w1w2, where A
?
=) w1 and B
?
=) w2.
Suppose w1 and w2 both have length at least 2. Construct within P an edge
labelled by uB going from the vertex where w1 ends to the vertex where w1
begins. By Lemma 5.1 uB =G w2. w1w2 =G 1, so uB =G w
1
1 . We have
now broken P down into two polygons with fewer than n sides, one labelled
w1uB, the other labelled u
1
b w2.
1
w1
w2 uB
w2
u−1B
1
w1
uB
1
Figure 4: Addition of an edge uB to triangulate a closed path w1w2
13
G. Now since G is a group, we have u = v.
From this result, a theorem about the closed paths in the Cayley graph can
be proved.
Theorem 5.2 If a nitely generated group G = hX ;Ri is context-free,
there exists a constant K such that every closed path in the Cayley graph
(G) can be K-triangulated. [12]
Proof: Let G be context-free, and let C be a reduced grammar in Chomsky
normal form which generates WP (G).
Let be a closed path in (G), labelled by w = y1; : : : ; yn, enclosing an
n-gon P . We will create a diagonal triangulation of P by adding edges
labelled by words (not letters) in X joining vertices of P so that the label
of the boundaries of any resulting polygons are true relations in G. The
result is trivial if n = 3, so assume n 4. If A is a variable of C, let uA
denote the shortest word derivable from A. Consider any derivation S
?
=) w.
Since C is in Chomsky normal form, this derivation must be of the form
S ) AB
?
=) w1w2, where A
?
=) w1 and B
?
=) w2.
Suppose w1 and w2 both have length at least 2. Construct within P an edge
labelled by uB going from the vertex where w1 ends to the vertex where w1
begins. By Lemma 5.1 uB =G w2. w1w2 =G 1, so uB =G w
1
1 . We have
now broken P down into two polygons with fewer than n sides, one labelled
w1uB, the other labelled u
1
b w2.
1
w1
w2 uB
w2
u−1B
1
w1
uB
1
Figure 4: Addition of an edge uB to triangulate a closed path w1w2
13
Page 14
Next suppose instead that one of w1 or w2 is a terminal symbol a. Say it's
w1. So S
?
=) aw2 and jw2j 3. Then there is a derivation S
?
=) aB )
aCD
?
=) aw21w22 where at least one of w21 or w22, say w22, has length 2.
Add an edge to P labelled by uD from the vertex where w22 begins to the
vertex where w22 ends. So now we have split P into two polygons, one
labelled by aw21uD and the other labelled by u
1
D w22.
a
w2
1 a
w21
w22 uD
1
a
w21
uD
1
u−1Dw22
1
Figure 5: Addition of an edge uB to triangulate a closed path aw2 = aw21w22
Iterating this procedure on each of the resulting smaller polygons will yield
a diagonal triangulation of P . Each edge of the triangulation is labelled by
a uA for some variable A from C, which has nitely many variables. So let
K = maxA2C juAj and we have a K-triangulation of P .
Next we will use the theory of ends to prove some results about the structure
of (G), and hence about the construction of G.
6 The Theory of Ends
Let be a graph with origin vertex v0. De ne (n) to be the set of points
of connected to v0 by a path of length no more than n.
n (n) is with all the points no more than n edges away from the origin
removed. This may break the graph into several separate components. An
end is, roughly speaking, a component that is left as n tends to in nity. For
example, the in nite cyclic group has two ends, and the free group of rank
2 has in nitely many ends.
14
w1. So S
?
=) aw2 and jw2j 3. Then there is a derivation S
?
=) aB )
aCD
?
=) aw21w22 where at least one of w21 or w22, say w22, has length 2.
Add an edge to P labelled by uD from the vertex where w22 begins to the
vertex where w22 ends. So now we have split P into two polygons, one
labelled by aw21uD and the other labelled by u
1
D w22.
a
w2
1 a
w21
w22 uD
1
a
w21
uD
1
u−1Dw22
1
Figure 5: Addition of an edge uB to triangulate a closed path aw2 = aw21w22
Iterating this procedure on each of the resulting smaller polygons will yield
a diagonal triangulation of P . Each edge of the triangulation is labelled by
a uA for some variable A from C, which has nitely many variables. So let
K = maxA2C juAj and we have a K-triangulation of P .
Next we will use the theory of ends to prove some results about the structure
of (G), and hence about the construction of G.
6 The Theory of Ends
Let be a graph with origin vertex v0. De ne (n) to be the set of points
of connected to v0 by a path of length no more than n.
n (n) is with all the points no more than n edges away from the origin
removed. This may break the graph into several separate components. An
end is, roughly speaking, a component that is left as n tends to in nity. For
example, the in nite cyclic group has two ends, and the free group of rank
2 has in nitely many ends.
14
Page 15
Formally de ne the number of ends of , e( ), by
e( ) = lim
n!1
(the number of in nite components of n (n)):
If G = hX ;Ri is a nitely generated group, we say that e(G) = e( (G)) for
brevity.
1
x
x−1
y
y−1
Γ Γ/Γ(1) Γ/Γ(2)
Figure 6: The Cayley graph (G) of the free group of order 2, G = hx; y ; ;i,
and with the components (1) and (2) removed.
Clearly the number of ends of a graph depends on how connected its vertices
are, so we shall prove a small result stating that, given a diagonal triangula-
tion of a polygon P whose boundary is divided into three consecutive arcs,
there is a triangle with at least one vertex on each arc.
Lemma 6.1 Let T be a diagonal triangulation of a polygon P with at least 3
edges. If the boundary edges of P are divided into three consecutive nonempty
arcs, then some triangle has vertices on all three arcs. Furthermore, this
triangle is unique. [12]
Proof: Denote the boundary of P by P . A triangle is said to be on P if
one or more of its edges are in P . A triangle is called critical if it has two
edges on P . Note that if T has more than one triangle, there are at least
two critical triangles. If all of the triangles in T with edges in P are critical,
the result holds. So assume there is at least one triangle t with only one edge
e in P and write P = 1e2, t = ee1e2. Then T gives a triangulation T1
of the polygon P1 bounded by 1ee1 and T2 of P2 bounded by 2e2e. Both
T1 and T2 have more than one triangle and thus have at least two critical
triangles, one of which may be t. The result follows by induction.
Now we will prove that the triangle is unique, by induction on the number
k of triangles in T and by 'colouring' the vertices of P . Assign a colour to
the vertices lying on each arc, so that three colours are used and the vertices
of each colour occur consecutively. If k = 1, the result is clearly true. If
15
e( ) = lim
n!1
(the number of in nite components of n (n)):
If G = hX ;Ri is a nitely generated group, we say that e(G) = e( (G)) for
brevity.
1
x
x−1
y
y−1
Γ Γ/Γ(1) Γ/Γ(2)
Figure 6: The Cayley graph (G) of the free group of order 2, G = hx; y ; ;i,
and with the components (1) and (2) removed.
Clearly the number of ends of a graph depends on how connected its vertices
are, so we shall prove a small result stating that, given a diagonal triangula-
tion of a polygon P whose boundary is divided into three consecutive arcs,
there is a triangle with at least one vertex on each arc.
Lemma 6.1 Let T be a diagonal triangulation of a polygon P with at least 3
edges. If the boundary edges of P are divided into three consecutive nonempty
arcs, then some triangle has vertices on all three arcs. Furthermore, this
triangle is unique. [12]
Proof: Denote the boundary of P by P . A triangle is said to be on P if
one or more of its edges are in P . A triangle is called critical if it has two
edges on P . Note that if T has more than one triangle, there are at least
two critical triangles. If all of the triangles in T with edges in P are critical,
the result holds. So assume there is at least one triangle t with only one edge
e in P and write P = 1e2, t = ee1e2. Then T gives a triangulation T1
of the polygon P1 bounded by 1ee1 and T2 of P2 bounded by 2e2e. Both
T1 and T2 have more than one triangle and thus have at least two critical
triangles, one of which may be t. The result follows by induction.
Now we will prove that the triangle is unique, by induction on the number
k of triangles in T and by 'colouring' the vertices of P . Assign a colour to
the vertices lying on each arc, so that three colours are used and the vertices
of each colour occur consecutively. If k = 1, the result is clearly true. If
15
Page 16
(a) (b)
η1
e
η2 e1 e2
(c)
Figure 7: A polygon P (a) triangulated so every triangle on P is critical,
(b) with two triangles on P not critical, (c) with one non-critical triangle
selected to break P into the smaller polygons 1ee1 and 2e2e.
Figure 8: A critical triangle t in the triangulation of a polygon P has vertices
of only two colours. The edges of t on P are removed to create a smaller
polygon P 0 with fewer triangles.
k > 1, let t be a critical triangle. If t has a vertex of each colour, then it
is the desired triangle and is unique, since if there was another triangle it
would have to intersect t. If t has vertices of only one or two colours, create
another polygon P 0 by deleting the edges of t which are on P . Then P 0
still has vertices of three colours (since each arc was on at least two vertices)
and its triangulation has fewer than k triangles so the result follows.
Now we can prove that context-free groups have more than one end, with
the aim of eventually showing this implies the group is constructed in a
certain way.
Lemma 6.2 If G is an in nite context-free group, G has more than one
end. [12]
Proof: Let = (G). If u; v are vertices in then de ne d(u; v) to be the
length of the shortest path between u and v. Choose the identity element 1
of G to be the origin of . G is in nite so there are arbitrarily long words
16
η1
e
η2 e1 e2
(c)
Figure 7: A polygon P (a) triangulated so every triangle on P is critical,
(b) with two triangles on P not critical, (c) with one non-critical triangle
selected to break P into the smaller polygons 1ee1 and 2e2e.
Figure 8: A critical triangle t in the triangulation of a polygon P has vertices
of only two colours. The edges of t on P are removed to create a smaller
polygon P 0 with fewer triangles.
k > 1, let t be a critical triangle. If t has a vertex of each colour, then it
is the desired triangle and is unique, since if there was another triangle it
would have to intersect t. If t has vertices of only one or two colours, create
another polygon P 0 by deleting the edges of t which are on P . Then P 0
still has vertices of three colours (since each arc was on at least two vertices)
and its triangulation has fewer than k triangles so the result follows.
Now we can prove that context-free groups have more than one end, with
the aim of eventually showing this implies the group is constructed in a
certain way.
Lemma 6.2 If G is an in nite context-free group, G has more than one
end. [12]
Proof: Let = (G). If u; v are vertices in then de ne d(u; v) to be the
length of the shortest path between u and v. Choose the identity element 1
of G to be the origin of . G is in nite so there are arbitrarily long words
16
Page 17
y1 : : : yj such that the shortest word equivalent to them is of length j. We
can translate the mid point of the path with label y1 : : : yj to the origin of
. This means that for any i, 9ui; vi such that d(ui; 1) = d(vi; 1) = i and
d(ui; vi) = 2i.
G is context-free so 9K such that every closed path in can beK-triangulated.
Pick n > 32K. Is it true that if i n then ui and vi are in di erent compo-
nents of n (n)? We will prove this by contradiction. Suppose that ui and
vi are in the same component of n (n). Let be a path of minimal length
from 1 to ui. Let
be a path of minimal length from vi to 1. Let be a
path in n (n) from ui to vi. By Lemma 6.1 some triangle t has vertices
a; b; c on ; ;
respectively.
α
β
γ
1
ui vi
a
b
c
Each edge of t represents a path of length no greater than K. Since b 2
n (n) we have d(1; a) n K, or we would get a path shorter than n to b
by going from 1 to a and then from a to b in a path of no longer than K steps.
So we now get that d(a; ui) i n+K since we have d(1; ui) = i and a lies
on a minimal path from 1 to ui. Similarly d(c; vi) i n+K. But now there
is a path ui ! a! c! vi of length 2i 2n+ 2K +K = 2i+ (3K 2n).
But this length is less than 2i since 2n > 3K. This contradicts d(ui; vi) = 2i,
so ui and vi are not in the same component of n (n).
Now that we know a context-free group has more than one end, we can use
Stallings' structure theorem to prove it has a certain construction in terms
of free products.
7 Torsion-free groups and free products
Let G be a group. If an element g of G has nite order, that is, there is
some nite n such that gn = 1, then g is a torsion element of G. If the only
torsion element in G is the identity, then G is torsion-free.
If G = hX ;Ri and H = hY ; Si are groups with X and Y disjoint sets, then
17
can translate the mid point of the path with label y1 : : : yj to the origin of
. This means that for any i, 9ui; vi such that d(ui; 1) = d(vi; 1) = i and
d(ui; vi) = 2i.
G is context-free so 9K such that every closed path in can beK-triangulated.
Pick n > 32K. Is it true that if i n then ui and vi are in di erent compo-
nents of n (n)? We will prove this by contradiction. Suppose that ui and
vi are in the same component of n (n). Let be a path of minimal length
from 1 to ui. Let
be a path of minimal length from vi to 1. Let be a
path in n (n) from ui to vi. By Lemma 6.1 some triangle t has vertices
a; b; c on ; ;
respectively.
α
β
γ
1
ui vi
a
b
c
Each edge of t represents a path of length no greater than K. Since b 2
n (n) we have d(1; a) n K, or we would get a path shorter than n to b
by going from 1 to a and then from a to b in a path of no longer than K steps.
So we now get that d(a; ui) i n+K since we have d(1; ui) = i and a lies
on a minimal path from 1 to ui. Similarly d(c; vi) i n+K. But now there
is a path ui ! a! c! vi of length 2i 2n+ 2K +K = 2i+ (3K 2n).
But this length is less than 2i since 2n > 3K. This contradicts d(ui; vi) = 2i,
so ui and vi are not in the same component of n (n).
Now that we know a context-free group has more than one end, we can use
Stallings' structure theorem to prove it has a certain construction in terms
of free products.
7 Torsion-free groups and free products
Let G be a group. If an element g of G has nite order, that is, there is
some nite n such that gn = 1, then g is a torsion element of G. If the only
torsion element in G is the identity, then G is torsion-free.
If G = hX ;Ri and H = hY ; Si are groups with X and Y disjoint sets, then
17
Page 18
the free product of G and H is de ned to be G?H = hX [Y ;R[Si. G and
H are called the factors of the free product. A free product is nontrivial if
neither of the factors is the trivial group.
If G is a group, the rank r(G) of G is the minimal number of generators of
G. A theorem of Grusko [6] states that r(G ? H) = r(G) + r(H).
Let G = hX ;Ri and H = hY ; Si be groups, with A and B respective
subgroups. If we have an isomorphism : A ! B, then the free product of
G and H amalgamating A and B is the group hX[Y ;R[S[fa = (a); a 2
Agi.
If G = hX ;Ri is a group, with A and B subgroups of G, and : A ! B
an isomorphism, then the HNN extension [9] of G with stable letter t and
associated subgroups A and B is the group hX [ftg ;R[ft 1at = (a); a 2
Agi.
Stallings' structure theorem [14] says that in the case that a group G is
nitely generated and torsion free, G has more than one end if and only if
G is the in nite cyclic group or is a nontrivial free product.
Theorem 7.1 A nitely generated torsion-free group is free if and only if
it is context-free. [12]
Proof: Let G be a context-free group. If r(G) = 0 then G is the trivial
group, which is free. If r(G) = 1 then G is the in nite cyclic group since
G is torsion-free. So suppose r(G) 2. G is in nite since it is a nontrivial
torsion-free group. By Lemma 6.2, G as more than one end so by Stallings'
structure theorem G is a nontrivial free product G = G1 ?G2. By Grushko's
theorem G1 and G2 have rank less than r(G). By Lemma 4.4 G1 and G2 are
context-free. Hence, G1 and G2 are free by the induction hypothesis. Since
the product of free groups is free, G is free.
Stallings' structure theorem can be stated more generally.
Theorem 7.2 (Stallings' structure theorem) A nitely generated group G
has more than one end if and only if G is eiter a nontrivial free product with
amalgamation or an HNN extension, where the amalgamated or associated
subgroups are nite. [14]
If G is a group, we say that G G0 G1 : : : Gn is an accessible series
for G if each Gi is a free product with amalgamation or an HNN extension
of Gi+1 and the amalgamated or associated subgroups are nite.
18
H are called the factors of the free product. A free product is nontrivial if
neither of the factors is the trivial group.
If G is a group, the rank r(G) of G is the minimal number of generators of
G. A theorem of Grusko [6] states that r(G ? H) = r(G) + r(H).
Let G = hX ;Ri and H = hY ; Si be groups, with A and B respective
subgroups. If we have an isomorphism : A ! B, then the free product of
G and H amalgamating A and B is the group hX[Y ;R[S[fa = (a); a 2
Agi.
If G = hX ;Ri is a group, with A and B subgroups of G, and : A ! B
an isomorphism, then the HNN extension [9] of G with stable letter t and
associated subgroups A and B is the group hX [ftg ;R[ft 1at = (a); a 2
Agi.
Stallings' structure theorem [14] says that in the case that a group G is
nitely generated and torsion free, G has more than one end if and only if
G is the in nite cyclic group or is a nontrivial free product.
Theorem 7.1 A nitely generated torsion-free group is free if and only if
it is context-free. [12]
Proof: Let G be a context-free group. If r(G) = 0 then G is the trivial
group, which is free. If r(G) = 1 then G is the in nite cyclic group since
G is torsion-free. So suppose r(G) 2. G is in nite since it is a nontrivial
torsion-free group. By Lemma 6.2, G as more than one end so by Stallings'
structure theorem G is a nontrivial free product G = G1 ?G2. By Grushko's
theorem G1 and G2 have rank less than r(G). By Lemma 4.4 G1 and G2 are
context-free. Hence, G1 and G2 are free by the induction hypothesis. Since
the product of free groups is free, G is free.
Stallings' structure theorem can be stated more generally.
Theorem 7.2 (Stallings' structure theorem) A nitely generated group G
has more than one end if and only if G is eiter a nontrivial free product with
amalgamation or an HNN extension, where the amalgamated or associated
subgroups are nite. [14]
If G is a group, we say that G G0 G1 : : : Gn is an accessible series
for G if each Gi is a free product with amalgamation or an HNN extension
of Gi+1 and the amalgamated or associated subgroups are nite.
18
Page 19
G is accessible if there is an upper bound on the lengths of the accessi-
ble series for G, called s, the accessibility length of G. Finite groups have
accessibility length zero since free products with amalgamation or HNN ex-
tensions are in nite. It follows from Grushko's theorem that torsion free
groups are accessible. In fact it has been proven by Dunwoody [2] that all
nitely presented groups are accessible.
Now we can nally prove our theorem! If the word problem of a group can
be solved on a pushdown automaton, then it is context free. So we need to
show that if a nitely generated group is context free, then it is virtually
free.
Theorem 7.3 Let G be a nitely generated context-free, accessible group.
Then G is virtually free.
Proof: We shall prove the result by induction on the accessibility length
s of G. If s = 0, G is nite and the result holds. So suppose that G =
hG1 ? G01 ; F = (F )i or G = hG1; t ; tF t
1 = (F )i with F a nite group.
G1 and G01 each have accessibility length at most s 1. By Lemma 4.4, G1
and G01 are context-free since they are subgroups of G, so they are virtually
free by the induction hypothesis.
Now we use results from Gregorac, Kerrass et al [5] which say that the class
of nitely generated virtually free groups is closed under free production
with amalgamation or HNN extension, where the associated or amalgamated
subgroups are nite. Thus G is virtually free.
8 One-counter automata
In the special case of a push-down automaton (Q;; ; ; i; F ) with = fgg,
the automaton is called a one-counter automaton [7, 10] because, since the
stack will always be of the form gn, it is sucient just to keep track of the
stack height, n.
A language accepted by a one-counter automaton is called a one-counter
language. It can be quite neatly shown that groups with one-counter word
problems are virtually cyclic, meaning they have a cyclic subgroup of nite
index. This was rst proved by Herbst [7], but we shall present here the
proof of Holt, Owen and Thomas [10].
First, we de ne the growth function
G(n) of a group G = hX ;Ri to be the
number of elements of G that are represented by words in X+ of length at
most n.
19
ble series for G, called s, the accessibility length of G. Finite groups have
accessibility length zero since free products with amalgamation or HNN ex-
tensions are in nite. It follows from Grushko's theorem that torsion free
groups are accessible. In fact it has been proven by Dunwoody [2] that all
nitely presented groups are accessible.
Now we can nally prove our theorem! If the word problem of a group can
be solved on a pushdown automaton, then it is context free. So we need to
show that if a nitely generated group is context free, then it is virtually
free.
Theorem 7.3 Let G be a nitely generated context-free, accessible group.
Then G is virtually free.
Proof: We shall prove the result by induction on the accessibility length
s of G. If s = 0, G is nite and the result holds. So suppose that G =
hG1 ? G01 ; F = (F )i or G = hG1; t ; tF t
1 = (F )i with F a nite group.
G1 and G01 each have accessibility length at most s 1. By Lemma 4.4, G1
and G01 are context-free since they are subgroups of G, so they are virtually
free by the induction hypothesis.
Now we use results from Gregorac, Kerrass et al [5] which say that the class
of nitely generated virtually free groups is closed under free production
with amalgamation or HNN extension, where the associated or amalgamated
subgroups are nite. Thus G is virtually free.
8 One-counter automata
In the special case of a push-down automaton (Q;; ; ; i; F ) with = fgg,
the automaton is called a one-counter automaton [7, 10] because, since the
stack will always be of the form gn, it is sucient just to keep track of the
stack height, n.
A language accepted by a one-counter automaton is called a one-counter
language. It can be quite neatly shown that groups with one-counter word
problems are virtually cyclic, meaning they have a cyclic subgroup of nite
index. This was rst proved by Herbst [7], but we shall present here the
proof of Holt, Owen and Thomas [10].
First, we de ne the growth function
G(n) of a group G = hX ;Ri to be the
number of elements of G that are represented by words in X+ of length at
most n.
19
Page 20
Theorem 8.1 If the word problem of a nitely generated group G is a one-
counter language, then S has a linear growth function. [10]
Proof: Let G = hX ;Ri be a nitely generated group. Let q be the number
of states of a one-counter automaton M accepting WP (G).
For each word w 2 X+, choose the shortest path p(w) in M that accepts
wxw 1, for some letter x. We need to show that there is a constant K such
that immediately after reading x in p(w), the stack height h(w) is at most
Kjwj. Then, after reading x in p(w), for words up to length n there are
only (Kn + 1)q possibilities for the pair (h(w); t), where t is the state of
the machine. This pair cannot be the same for two words w1 and w2 that
represent di erent group elements because otherwise there would be a path
w1xw
1
2 accepted by the automaton, implying that w1 and w2 do in fact
represent the same element.
To show that h(w) Kjwj, rst assume that all moves in M change the
stack height by at most one. Allow M , without loss of generality, to only
make reading moves, where it reads from the input and changes state, or a
non-reading move, where it changes the stack and changes state. So only
non-reading moves can change the stack height.
If h(w) > q(n+ 1) then, at some point reading w and between two reading
moves, there must be an occasion when the stack height increases by at least
q. Clearly M must repeat states at this point, so there is a loop in p(w)
linking the repeated states during which the stack height is increased by r,
with 0 < r q. Similarly, when reading w 1, there must be a gap when the
stack height is reduced by some u, 0 < u q.
If h(w) > q3(n+ 1) then we can nd gaps containing q2 loops of this kind,
in which the stack height is increased by at most q when reading w and
similarly when reading w 1 there are gaps containing q2 loops in which the
stack height is decreased by at most q.
Amongst the increasing loops, at least q of them must increase the stack
height by the same number r q, and there are similarly at least q decreas-
ing loops which decrease the stack height by some u q.
If we remove u loops which increase the stack height by r, and r loops
which decrease the stack height by u, we make a shorter path which accepts
wxw 1, contradicting the minimality of p(w). We can only do this if doing
so does not cause the stack to be empty at any stage.
Assume that h(w) > q3(n + 2). Choose the gap between reading-moves in
which we remove the increasing loops to be the latest one in which the stack
height increases by at least q3 at some stage during the gap, and remove
20
counter language, then S has a linear growth function. [10]
Proof: Let G = hX ;Ri be a nitely generated group. Let q be the number
of states of a one-counter automaton M accepting WP (G).
For each word w 2 X+, choose the shortest path p(w) in M that accepts
wxw 1, for some letter x. We need to show that there is a constant K such
that immediately after reading x in p(w), the stack height h(w) is at most
Kjwj. Then, after reading x in p(w), for words up to length n there are
only (Kn + 1)q possibilities for the pair (h(w); t), where t is the state of
the machine. This pair cannot be the same for two words w1 and w2 that
represent di erent group elements because otherwise there would be a path
w1xw
1
2 accepted by the automaton, implying that w1 and w2 do in fact
represent the same element.
To show that h(w) Kjwj, rst assume that all moves in M change the
stack height by at most one. Allow M , without loss of generality, to only
make reading moves, where it reads from the input and changes state, or a
non-reading move, where it changes the stack and changes state. So only
non-reading moves can change the stack height.
If h(w) > q(n+ 1) then, at some point reading w and between two reading
moves, there must be an occasion when the stack height increases by at least
q. Clearly M must repeat states at this point, so there is a loop in p(w)
linking the repeated states during which the stack height is increased by r,
with 0 < r q. Similarly, when reading w 1, there must be a gap when the
stack height is reduced by some u, 0 < u q.
If h(w) > q3(n+ 1) then we can nd gaps containing q2 loops of this kind,
in which the stack height is increased by at most q when reading w and
similarly when reading w 1 there are gaps containing q2 loops in which the
stack height is decreased by at most q.
Amongst the increasing loops, at least q of them must increase the stack
height by the same number r q, and there are similarly at least q decreas-
ing loops which decrease the stack height by some u q.
If we remove u loops which increase the stack height by r, and r loops
which decrease the stack height by u, we make a shorter path which accepts
wxw 1, contradicting the minimality of p(w). We can only do this if doing
so does not cause the stack to be empty at any stage.
Assume that h(w) > q3(n + 2). Choose the gap between reading-moves in
which we remove the increasing loops to be the latest one in which the stack
height increases by at least q3 at some stage during the gap, and remove
20
Page 21
loops as late as possible during that gap. Similarly, choose the earliest
possible gap to remove decreasing loops. Between the increasing gap and
the end of reading w, the stack cannot be empty because then there would
be a later gap which increases the stack height by q3, and a similar argument
follows for the decreasing gaps.
Between the beginning of reading w 1 and the place where we removed the
decreasing loops, the stack height decreases by less than q3 during each gap,
and hence less than q3(n + 1) in total. Since we are not removing all the
loops after this gap, the stack height could decrease by some number less
than q3 during the rest of p(w). But since h(w) > q3(n+ 2), this means the
stack can never be empty. By this contradiction, we get h(w) q3(n + 2),
proving the result.
Now we need the concept of ShortLex ordering. Suppose X is a nite alpha-
bet with a linear order <X on it. Then if ; 2 X?, <SL if either:
j j < j j or
a1a2 : : : am , b1b2 : : : bm and 9k with 1 k m such that
a1 = b1, a2 = b2 , : : : , ak 1 = bk 1 , ak <X bk.
Call the (unique) least representative of a group element under <SL the
ShortLex normal form of that element.
Note that if uvw is in ShortLex normal form, so is v, since if there is a
v0 <SL v, then uv0w <SL uvw. This means that any subword of a word in
ShortLex normal form is itself in ShortLex normal form. By convention, the
empty word is not in ShortLex normal form.
Theorem 8.2 If G is a group with linear growth function then there exist
elements ai; bj ; ck 2 G , with 1 i; j; k N for some N , such that any
element of G has the form aibni ci for some i and n with 1 i N and
n 0. [10]
Proof: Let G = hX ;Ri be a nitely generated group with linear growth.
Choose a linear order on X and then consider the set N of ShortLex normal
forms for G.
Let L N denote the set of words w 2 N such that w is a pre x of in nitely
many v 2 N , i.e. w can be extended inde nitely.
If G is nite then the result is trivial, so assume G is in nite, and hence L
is in nite since every element has a distinct ShortLex normal form.
21
possible gap to remove decreasing loops. Between the increasing gap and
the end of reading w, the stack cannot be empty because then there would
be a later gap which increases the stack height by q3, and a similar argument
follows for the decreasing gaps.
Between the beginning of reading w 1 and the place where we removed the
decreasing loops, the stack height decreases by less than q3 during each gap,
and hence less than q3(n + 1) in total. Since we are not removing all the
loops after this gap, the stack height could decrease by some number less
than q3 during the rest of p(w). But since h(w) > q3(n+ 2), this means the
stack can never be empty. By this contradiction, we get h(w) q3(n + 2),
proving the result.
Now we need the concept of ShortLex ordering. Suppose X is a nite alpha-
bet with a linear order <X on it. Then if ; 2 X?, <SL if either:
j j < j j or
a1a2 : : : am , b1b2 : : : bm and 9k with 1 k m such that
a1 = b1, a2 = b2 , : : : , ak 1 = bk 1 , ak <X bk.
Call the (unique) least representative of a group element under <SL the
ShortLex normal form of that element.
Note that if uvw is in ShortLex normal form, so is v, since if there is a
v0 <SL v, then uv0w <SL uvw. This means that any subword of a word in
ShortLex normal form is itself in ShortLex normal form. By convention, the
empty word is not in ShortLex normal form.
Theorem 8.2 If G is a group with linear growth function then there exist
elements ai; bj ; ck 2 G , with 1 i; j; k N for some N , such that any
element of G has the form aibni ci for some i and n with 1 i N and
n 0. [10]
Proof: Let G = hX ;Ri be a nitely generated group with linear growth.
Choose a linear order on X and then consider the set N of ShortLex normal
forms for G.
Let L N denote the set of words w 2 N such that w is a pre x of in nitely
many v 2 N , i.e. w can be extended inde nitely.
If G is nite then the result is trivial, so assume G is in nite, and hence L
is in nite since every element has a distinct ShortLex normal form.
21
Page 22
Form a graph with vertex set L and edges from each w to wx for each
x 2 X such that wa 2 L. Add an origin vertex to the graph, with edges
labelled by x leading from it to the vertex representing the word x, when
x 2 L. This graph is a subgraph of the Cayley graph of G, with the property
that there is a unique path from the origin to each vertex, so is a tree.
Let K(n), n 0 be the number of words in L of length n, which is equal
to the number of vertices of at a distance n from the origin. Since is a
tree with no nite branches, every vertex at a distance n from the origin is
connected to at least one vertex at a distance n+1, so K(n) is an increasing
function of n. But G has linear growth function, and
G(n)
nX
i=1
K(i)
implies that K(n) is bounded. So 9K with K(n) = K for all suciently
large n.
So, once n is large enough that K(n) = K, consists of K disjoint paths,
or strands. Let w1; : : : ; wK be pre xes of the strands such that no wi is a
pre x of another wj . What this means in e ect is that each wi is picked
to be on the strand after the last 'branching' vertex which is connected to
more than one longer word. Note that every word in X? has a pre x one of
the wi.
We will now split up one particular strand, say that whose pre x is w1, into
segments p1n. Let p11 = w1. Then we can read further along the strand
to w , where 2 X? and has a pre x p12 which is one of the wi. We
can continue in this vein to make an in nite sequence fp1ng11 with each p1i
equal to some wk. Clearly p1i must repeat at some point, say p1i = p1j ,
where i < j. Then p1(i+k) = p1(j+k) for all k 0.
So the whole strand consists of p11p12 : : : p1(i 1) followed by in nitely many
repetitions of y1 := p1i : : : p1(j 1). To put it another way, for each i, the
in nite strand with pre x wi consists of a pre x followed by in nitely many
repetitions of yi. Let
B = fbj : bj is a cyclic permutation ofyi; 1 i Kg:
Then all words in L are of the form aibnj , where bj 2 B and the ai are nitely
many pre xes of the graph.
Now consider the graph 0 constructed from N in the same way was
constructed from L. 0 consists of with nite branches added to some
vertices. If the lengths of these branches are uniformly bounded, then their
vertices represent a nite set of elements of ck.
Suppose the branches are not uniformly bounded. Then there exist arbitrar-
ily long branches. So there exists a pre x u and some v 2 B, for which there
22
x 2 X such that wa 2 L. Add an origin vertex to the graph, with edges
labelled by x leading from it to the vertex representing the word x, when
x 2 L. This graph is a subgraph of the Cayley graph of G, with the property
that there is a unique path from the origin to each vertex, so is a tree.
Let K(n), n 0 be the number of words in L of length n, which is equal
to the number of vertices of at a distance n from the origin. Since is a
tree with no nite branches, every vertex at a distance n from the origin is
connected to at least one vertex at a distance n+1, so K(n) is an increasing
function of n. But G has linear growth function, and
G(n)
nX
i=1
K(i)
implies that K(n) is bounded. So 9K with K(n) = K for all suciently
large n.
So, once n is large enough that K(n) = K, consists of K disjoint paths,
or strands. Let w1; : : : ; wK be pre xes of the strands such that no wi is a
pre x of another wj . What this means in e ect is that each wi is picked
to be on the strand after the last 'branching' vertex which is connected to
more than one longer word. Note that every word in X? has a pre x one of
the wi.
We will now split up one particular strand, say that whose pre x is w1, into
segments p1n. Let p11 = w1. Then we can read further along the strand
to w , where 2 X? and has a pre x p12 which is one of the wi. We
can continue in this vein to make an in nite sequence fp1ng11 with each p1i
equal to some wk. Clearly p1i must repeat at some point, say p1i = p1j ,
where i < j. Then p1(i+k) = p1(j+k) for all k 0.
So the whole strand consists of p11p12 : : : p1(i 1) followed by in nitely many
repetitions of y1 := p1i : : : p1(j 1). To put it another way, for each i, the
in nite strand with pre x wi consists of a pre x followed by in nitely many
repetitions of yi. Let
B = fbj : bj is a cyclic permutation ofyi; 1 i Kg:
Then all words in L are of the form aibnj , where bj 2 B and the ai are nitely
many pre xes of the graph.
Now consider the graph 0 constructed from N in the same way was
constructed from L. 0 consists of with nite branches added to some
vertices. If the lengths of these branches are uniformly bounded, then their
vertices represent a nite set of elements of ck.
Suppose the branches are not uniformly bounded. Then there exist arbitrar-
ily long branches. So there exists a pre x u and some v 2 B, for which there
22
Page 23
exist arbitrarily long words w such that, for some m, uvmw 2 N but uvmw1
is not in nitely extensible, where w1 is the rst letter of w. In particular, w
does not have v as a pre x because otherwise we could pick a di erent w0
and consider uvm+1w0.
Suppose that the number of elements of G of length at most n is at most
Cn, which we can do since G has linear growth. As jwj increases, so must
m, so we can choose m and jwj suciently large that mjwj > C(mjvj+ jwj).
The mjwj subwords of vmw of the form v0w(t), for 1 i m, 1 t jwj,
where w(t) denotes the pre x of w of length t, cannot all represent distinct
elements. Since they are all in ShortLex normal form, two of them must
be the same word, viw(s) vjw(t). Clearly i 6= j, as that would give
w(s) = w(t), which implies that s = t. Thus w has v as a pre x. This
contradiction proves that the added branches are uniformly bounded, so
there are nitely many elements ck and any element of G has the form
aibni ci.
This leads quite quickly to the following theorem:
Theorem 8.3 A nitely generated group G has one-counter word problem
if and only if G is virtually cyclic. [7, 10]
Proof: Suppose that G has one-counter word problem. By 8.2 there exist
elements a1; : : : ; an; b1; : : : ; bn; c1; : : : ; cn 2 G such that
G = [ni=1 [
1
r=0 aib
r
i ci
Given this, we have,
G = [ni=1(aihbiia
1
i )aici:
So G is a union of nitely many cosets. At least one of the subgroups
aihbiia
1
i has nite index in G, so G is virtually cyclic.
9 A characterisation of nitely generated groups
with soluble word problem
Now we will take a step away from the properties of a group's word problem
and instead consider what having a soluble word problem says about the
structure of a group.
23
is not in nitely extensible, where w1 is the rst letter of w. In particular, w
does not have v as a pre x because otherwise we could pick a di erent w0
and consider uvm+1w0.
Suppose that the number of elements of G of length at most n is at most
Cn, which we can do since G has linear growth. As jwj increases, so must
m, so we can choose m and jwj suciently large that mjwj > C(mjvj+ jwj).
The mjwj subwords of vmw of the form v0w(t), for 1 i m, 1 t jwj,
where w(t) denotes the pre x of w of length t, cannot all represent distinct
elements. Since they are all in ShortLex normal form, two of them must
be the same word, viw(s) vjw(t). Clearly i 6= j, as that would give
w(s) = w(t), which implies that s = t. Thus w has v as a pre x. This
contradiction proves that the added branches are uniformly bounded, so
there are nitely many elements ck and any element of G has the form
aibni ci.
This leads quite quickly to the following theorem:
Theorem 8.3 A nitely generated group G has one-counter word problem
if and only if G is virtually cyclic. [7, 10]
Proof: Suppose that G has one-counter word problem. By 8.2 there exist
elements a1; : : : ; an; b1; : : : ; bn; c1; : : : ; cn 2 G such that
G = [ni=1 [
1
r=0 aib
r
i ci
Given this, we have,
G = [ni=1(aihbiia
1
i )aici:
So G is a union of nitely many cosets. At least one of the subgroups
aihbiia
1
i has nite index in G, so G is virtually cyclic.
9 A characterisation of nitely generated groups
with soluble word problem
Now we will take a step away from the properties of a group's word problem
and instead consider what having a soluble word problem says about the
structure of a group.
23
Page 24
We will show a famous result of Boone and Higman [3], that a nitely-
generated group G has soluble word problem if and only if there exist a
simple group H and a nitely-presented group K such that G is a subgroup
of H and H is a subgroup of K. First, we will need to de ne some more
terms.
A group G is called simple if the only normal subgroups it contains are the
trivial group and G itself.
A group G = hX ;Ri is called recursively enumerable if there is a recursive
function which produces all of the words in X?. If a group is recursively
enumerable we can order its words, and we will refer to the ith word in this
ordering as wi.
Given a group G = hX ;Ri, we form the group Gy by a sequence of HNN
extensions. First, we add a generator t and no relations to obtain the free
product G ? hti. Then, we add generators ui and relations of the form
u 1i tui = twi for each of the words wi in the canonical ordering of X
?.
Finally, we add generators vi and relations of the form v
1
i tvi = t
1w 1i twi,
for each wi. We call the letters t, ui and vi the stable letters of Gy.
We will denote application of the y extension n times on a group by G(n),
and let G(0) = G. Let G(1) =
S1
n=1G
(n). The rank of a word w 2 G(1) is
the smallest n such that w 2 G(n).
Given a group G = hX ;Ri and any word w of G, Gw is the group obtained
by adding the relation w = 1 to G, that is, hX ;R [ fw = 1gi.
Lemma 9.1 Let G? be an HNN extension of the group G with stable letters
pv. Then G G?. [4]
Corollary 9.2 For any group G = hX ;Ri, G is embedded in Gy by the
identity map. Furthermore, given non-negative integers m and n such that
m n, the group G(m) is embedded in the group G(n) via the identity map.
[3]
Proof: It is clear that G Gy by the construction of Gy and Lemma 9.1.
It is also clear by induction on n m that G(m) G(n) when m n.
Given two decision problems (for example the word problem) P1 and P2, P1
is said to be uniformly soluble in P2 if a machine which solves P2 can also
solve P1. Our next step is to show that the word problem of Gy is uniformly
soluble in the word problem of G, and hence that the word problem of G(1)
is uniformly soluble in WP (G) as well.
24
generated group G has soluble word problem if and only if there exist a
simple group H and a nitely-presented group K such that G is a subgroup
of H and H is a subgroup of K. First, we will need to de ne some more
terms.
A group G is called simple if the only normal subgroups it contains are the
trivial group and G itself.
A group G = hX ;Ri is called recursively enumerable if there is a recursive
function which produces all of the words in X?. If a group is recursively
enumerable we can order its words, and we will refer to the ith word in this
ordering as wi.
Given a group G = hX ;Ri, we form the group Gy by a sequence of HNN
extensions. First, we add a generator t and no relations to obtain the free
product G ? hti. Then, we add generators ui and relations of the form
u 1i tui = twi for each of the words wi in the canonical ordering of X
?.
Finally, we add generators vi and relations of the form v
1
i tvi = t
1w 1i twi,
for each wi. We call the letters t, ui and vi the stable letters of Gy.
We will denote application of the y extension n times on a group by G(n),
and let G(0) = G. Let G(1) =
S1
n=1G
(n). The rank of a word w 2 G(1) is
the smallest n such that w 2 G(n).
Given a group G = hX ;Ri and any word w of G, Gw is the group obtained
by adding the relation w = 1 to G, that is, hX ;R [ fw = 1gi.
Lemma 9.1 Let G? be an HNN extension of the group G with stable letters
pv. Then G G?. [4]
Corollary 9.2 For any group G = hX ;Ri, G is embedded in Gy by the
identity map. Furthermore, given non-negative integers m and n such that
m n, the group G(m) is embedded in the group G(n) via the identity map.
[3]
Proof: It is clear that G Gy by the construction of Gy and Lemma 9.1.
It is also clear by induction on n m that G(m) G(n) when m n.
Given two decision problems (for example the word problem) P1 and P2, P1
is said to be uniformly soluble in P2 if a machine which solves P2 can also
solve P1. Our next step is to show that the word problem of Gy is uniformly
soluble in the word problem of G, and hence that the word problem of G(1)
is uniformly soluble in WP (G) as well.
24
Page 25
Lemma 9.3 For any group G, WP (Gy) is uniformly soluble in WP (G).
Furthermore, for any N 0, WP (G(N)) is uniformly soluble in WP (G).
[3]
We omit the proof, which can be found in [3].
Lemma 9.4 For any group G, suppose w = 1 in G(1). Then w = 1 in
G(n) where w is of rank n. [3]
Proof: The fact that w is trivial inG(1) is a result of nitely many relations
involving nitely many generators. Say these generators all fall in some
G(m), so that w = 1 in G(m). But n m and so by Lemma 9.2, w = 1 in
G(n).
Lemma 9.5 For any group G and any n 0, the group G(n) is embedded
in G(1) via the identity map. [3]
Proof: This is clear, by Lemma 9.2 and Lemma 9.4.
Proposition 9.6 For any group G, WP (G(1)) is uniformly soluble in WP (G).
[3]
Proof: By Lemmas 9.4 and 9.5, for any word w of G(1) of rank N , w = 1
in G(1) if and only if w = 1 in G(N). Now by Lemma 9.3, WP (G(N)) is
uniformly soluble in WP (G), so the result holds.
Now we will prove that G(1) is simple.
Lemma 9.7 [3] Let G = hX ;Ri be a group and w 2 X? be a word in G
such that w 6=G 1. Then 8w0 2 X?, w0 = 1 in Gyw.
Proof: Note that w = wi for some i in the enumeration of X?, and Gy
contains a relation v 1i tvi = t
1w 1i twi. Then since wi = w = 1 in G
yw, we
get that t = 1 in Gyw. Now, for every word w0 2 X?, rstly w0 = wj for
some j, and then there is a relation from the presentation of Gy of the form
u 1j tuj = twj , which gives us that wj = 1 in G
yw and so w0 = 1 in Gyw.
Lemma 9.8 Let G = hX ;Ri. G is simple if and only if for each w 2 X?
such that w 6=G 1, Gw is the trivial group.
25
Furthermore, for any N 0, WP (G(N)) is uniformly soluble in WP (G).
[3]
We omit the proof, which can be found in [3].
Lemma 9.4 For any group G, suppose w = 1 in G(1). Then w = 1 in
G(n) where w is of rank n. [3]
Proof: The fact that w is trivial inG(1) is a result of nitely many relations
involving nitely many generators. Say these generators all fall in some
G(m), so that w = 1 in G(m). But n m and so by Lemma 9.2, w = 1 in
G(n).
Lemma 9.5 For any group G and any n 0, the group G(n) is embedded
in G(1) via the identity map. [3]
Proof: This is clear, by Lemma 9.2 and Lemma 9.4.
Proposition 9.6 For any group G, WP (G(1)) is uniformly soluble in WP (G).
[3]
Proof: By Lemmas 9.4 and 9.5, for any word w of G(1) of rank N , w = 1
in G(1) if and only if w = 1 in G(N). Now by Lemma 9.3, WP (G(N)) is
uniformly soluble in WP (G), so the result holds.
Now we will prove that G(1) is simple.
Lemma 9.7 [3] Let G = hX ;Ri be a group and w 2 X? be a word in G
such that w 6=G 1. Then 8w0 2 X?, w0 = 1 in Gyw.
Proof: Note that w = wi for some i in the enumeration of X?, and Gy
contains a relation v 1i tvi = t
1w 1i twi. Then since wi = w = 1 in G
yw, we
get that t = 1 in Gyw. Now, for every word w0 2 X?, rstly w0 = wj for
some j, and then there is a relation from the presentation of Gy of the form
u 1j tuj = twj , which gives us that wj = 1 in G
yw and so w0 = 1 in Gyw.
Lemma 9.8 Let G = hX ;Ri. G is simple if and only if for each w 2 X?
such that w 6=G 1, Gw is the trivial group.
25
Page 26
Proof: Suppose there is some w such that Gw is not the trivial group, i.e.
it contains some nontrivial element n. Then fgng 1 : g 2 Gg is a normal
subgroup of G, so G is not simple.
Proposition 9.9 For any group G, the group G(1) is simple. [3]
Proof: Suppose w and v are words of G(1), and w 6= 1 in G(1). Let N be
the maximum of the ranks of w and v. By Lemma 9.5, w 6= 1 in G(N).
By Lemma 9.7, v = 1 in G(N+1)w. Since each generator and relation of
G(N+1)w is among those of G(1)w, v = 1 in G(1)w. Now G(1)w is the trivial
group 8w, so by Lemma 9.8 G(1) is simple.
We just need a couple more results from [8] and [9], respectively, and then
we will be able to prove our main theorem.
Lemma 9.10 A nitely generated group can be embedded in a nitely pre-
sented group if and only if it is recursively presented. [8]
Lemma 9.11 Any recursively generated group G can be embedded in a
group H generated by only two elements. [9]
Theorem 9.12 If a nitely generated group G has soluble word problem,
then there exist a simple group H and a nitely presented group K such that
G is a subgroup of H and H is a subgroup of K. [3]
Proof: Let G = hX ;Ri be a nitely generated group with soluble word
problem. We let H = G(1). Then by Proposition 9.9, H is simple.
But we can also give a presentation ofG(1) in terms of generators tn; ui;n; vi;n
and relations of the form w = 1 for each w 2WP (G(1)). By Proposition 9.6,
this set of relations is a recursive set.
Then, by Lemma 9.11, G(1) can be embedded in a group N consisting
of two generators and a resursively enumerable set of relations. Finally,
by Lemma 9.10, N can be embedded in a nitely presented group K, as
required.
We can also show the converse.
Theorem 9.13 Let G = hX ;Ri be a nitely generated group which is con-
tained in a simple group H, which is itself contained in a nitely presented
group K. Then G has soluble word problem. [3]
26
it contains some nontrivial element n. Then fgng 1 : g 2 Gg is a normal
subgroup of G, so G is not simple.
Proposition 9.9 For any group G, the group G(1) is simple. [3]
Proof: Suppose w and v are words of G(1), and w 6= 1 in G(1). Let N be
the maximum of the ranks of w and v. By Lemma 9.5, w 6= 1 in G(N).
By Lemma 9.7, v = 1 in G(N+1)w. Since each generator and relation of
G(N+1)w is among those of G(1)w, v = 1 in G(1)w. Now G(1)w is the trivial
group 8w, so by Lemma 9.8 G(1) is simple.
We just need a couple more results from [8] and [9], respectively, and then
we will be able to prove our main theorem.
Lemma 9.10 A nitely generated group can be embedded in a nitely pre-
sented group if and only if it is recursively presented. [8]
Lemma 9.11 Any recursively generated group G can be embedded in a
group H generated by only two elements. [9]
Theorem 9.12 If a nitely generated group G has soluble word problem,
then there exist a simple group H and a nitely presented group K such that
G is a subgroup of H and H is a subgroup of K. [3]
Proof: Let G = hX ;Ri be a nitely generated group with soluble word
problem. We let H = G(1). Then by Proposition 9.9, H is simple.
But we can also give a presentation ofG(1) in terms of generators tn; ui;n; vi;n
and relations of the form w = 1 for each w 2WP (G(1)). By Proposition 9.6,
this set of relations is a recursive set.
Then, by Lemma 9.11, G(1) can be embedded in a group N consisting
of two generators and a resursively enumerable set of relations. Finally,
by Lemma 9.10, N can be embedded in a nitely presented group K, as
required.
We can also show the converse.
Theorem 9.13 Let G = hX ;Ri be a nitely generated group which is con-
tained in a simple group H, which is itself contained in a nitely presented
group K. Then G has soluble word problem. [3]
26
Page 27
Proof: First note that there exists a recursive functional
which, when
applied to a nite presentation V of a group, yields a recursive enumeration
of all the consequent relations of V .
Suppose we have a nitely generated group G, contained in a simple group H
which is contained in a nitely presented group K. Note that if H is trivial
then so is G and thus WP (G) is soluble. So assume that H is non-trivial.
Suppose we have a word w 2 X? such that w =G 1. Then if we check the
enumeration
(K) of all the consequent relations of K, we will eventually
reach the one of the form w = 1. But if we are not sure that w is trivial
in G, we will not know when to stop looking, so we can not say for certain
that a word is not trivial in G. We might just need to look for a bit longer
and nd a relation that says it is trivial. So we need another process, which
we can perform in parallel with this one, which will de nitely tell us that a
non-trivial word is non-trivial in a nite amount of time.
Let wH0 be a word of H such that wH0 6=H 1. Let be the map embedding
G in H, and let be the map embedding H in K. is recursive,
meaning it can be computed, since its behaviour depends on the nitely
many generators of G. We have, for any word wH of H, with wH 6=H 1,
that
wH0 2 H = fwHgH ;
since H is simple, and that
(fwHgH) f (wH)gK
since H K.
Thus, for any w 2 X?, if w 6=G 1, then (w) 6=H 1 and (wH0) 2
f ( (w))gK , i.e (wH0) = 1 in the group K ( (w)) which is K with the
added relation psi( (w)) = 1.
Which is to say, if, where w 2 X?, (wH0) = 1 in K ( (w)), then w 6=G 1.
Now, suppose w is any word of G. First, compute w0 = ( (w)). Next,
apply
to K and K ( (w)). Now we can check through the relations in
(K) and
(K ( (w))) until we decide either that w0 = 1 in the rst set, in
which case w =G 1, or that (wH0) = 1 in the second set, implying that
w 6=G 1.
References
[1] A. V. Ansmov. The group languages. Kibernetika (Kiev), (4):18{24,
1971.
27
which, when
applied to a nite presentation V of a group, yields a recursive enumeration
of all the consequent relations of V .
Suppose we have a nitely generated group G, contained in a simple group H
which is contained in a nitely presented group K. Note that if H is trivial
then so is G and thus WP (G) is soluble. So assume that H is non-trivial.
Suppose we have a word w 2 X? such that w =G 1. Then if we check the
enumeration
(K) of all the consequent relations of K, we will eventually
reach the one of the form w = 1. But if we are not sure that w is trivial
in G, we will not know when to stop looking, so we can not say for certain
that a word is not trivial in G. We might just need to look for a bit longer
and nd a relation that says it is trivial. So we need another process, which
we can perform in parallel with this one, which will de nitely tell us that a
non-trivial word is non-trivial in a nite amount of time.
Let wH0 be a word of H such that wH0 6=H 1. Let be the map embedding
G in H, and let be the map embedding H in K. is recursive,
meaning it can be computed, since its behaviour depends on the nitely
many generators of G. We have, for any word wH of H, with wH 6=H 1,
that
wH0 2 H = fwHgH ;
since H is simple, and that
(fwHgH) f (wH)gK
since H K.
Thus, for any w 2 X?, if w 6=G 1, then (w) 6=H 1 and (wH0) 2
f ( (w))gK , i.e (wH0) = 1 in the group K ( (w)) which is K with the
added relation psi( (w)) = 1.
Which is to say, if, where w 2 X?, (wH0) = 1 in K ( (w)), then w 6=G 1.
Now, suppose w is any word of G. First, compute w0 = ( (w)). Next,
apply
to K and K ( (w)). Now we can check through the relations in
(K) and
(K ( (w))) until we decide either that w0 = 1 in the rst set, in
which case w =G 1, or that (wH0) = 1 in the second set, implying that
w 6=G 1.
References
[1] A. V. Ansmov. The group languages. Kibernetika (Kiev), (4):18{24,
1971.
27
Page 28
[2] C. Bamford and M. J. Dunwoody. On accessible groups. J. Pure Appl.
Algebra, 7(3):333{346, 1976.
[3] William W. Boone and Graham Higman. An algebraic characterization
of groups with soluble word problem. J. Austral. Math. Soc., 18:41{
53, 1974. Collection of articles dedicated to the memory of Hanna
Neumann, IX.
[4] John L. Britton. The word problem. Ann. of Math. (2), 77:16{32, 1963.
[5] R. Gregorac. On generalized free products of nite extensions of free
groups. J. London Math. Soc., 41:662{666, 1966.
[6] I. Grusko. O bazisah svobodnogo proizvedeniya grupp. Mat. Sbornik,
1940.
[7] Thomas Herbst. On a subclass of context-free groups. RAIRO Inform.
Theor. Appl., 25(3):255{272, 1991.
[8] G. Higman. Subgroups of nitely presented groups. Proc. Roy. Soc.
Ser. A, 262:455{475, 1961.
[9] Graham Higman, B. H. Neumann, and Hanna Neumann. Embedding
theorems for groups. J. London Math. Soc., 24:247{254, 1949.
[10] Owens Matthew D. Holt, Derek F. and Richard M. Thomas. Groups
and semigroups with a one-counter word problem.
[11] John E. Hopcroft and Je rey D. Ullman. Introduction to automata
theory, languages, and computation. Addison-Wesley Publishing Co.,
Reading, Mass., 1979. Addison-Wesley Series in Computer Science.
[12] David E. Muller and Paul E. Schupp. Groups, the theory of ends, and
context-free languages. J. Comput. System Sci., 26(3):295{310, 1983.
[13] Otto Schreier. Uber die Erweiterung von Gruppen I. Monatsh. Math.
Phys., 34(1):165{180, 1926.
[14] John R. Stallings. On torsion-free groups with in nitely many ends.
Ann. of Math. (2), 88:312{334, 1968.
28
Algebra, 7(3):333{346, 1976.
[3] William W. Boone and Graham Higman. An algebraic characterization
of groups with soluble word problem. J. Austral. Math. Soc., 18:41{
53, 1974. Collection of articles dedicated to the memory of Hanna
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28
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