Solution to the exercises

Abstract

a −1 b and ba −1 solve the two equations. Conversely, if ax = a has a solution, then there exists er such that aer = a; if b ∈ G, then b = ya, some y, and therefore ber = (ya) er = y(ae)r = ya = b; hence er is neutral on the right for all b ∈ G, and similarly there exists el, neutral on the left. Multiplication of the two gives el er = el, el er = er, and ii) follows. As for iii), solving ax = e and ya = e, if ax = az we have y(ax) = y(az) from which (ya)x = (ya)z, i.e. ex = ez and x = z; hence the right inverse is unique, and so is the left inverse. If ax = e, then xa = exa = yaxa = y(ax)a = yea = ya = e and ax = e = xa.