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 e _r such that ae _r = a; if b ∈ G, then b = ya, some y, and therefore be _r = (ya) e _r = y(ae) _r = ya = b; hence e _r is neutral on the right for all b ∈ G, and similarly there exists e _l , neutral on the left. Multiplication of the two gives e _l e _r = e _l , e _l e _r = e _r , 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.