The group \(\{ G,{\text{ }} * \} \) is Abelian and the bijection \(f:{\text{ }}G \to G\) is defined by \(f(x) = {x^{ - 1}},{\text{ }}x \in G\).

Show that \(f\) is an isomorphism.

we need to show that \(f(a * b) = f(a) * f(b)\)     R1

Note:     This R1 may be awarded at any stage.

let \(a,{\text{ }}b \in G\)     (M1)

consider \(f(a) * f(b)\)     M1

\( = {a^{ - 1}} * {b^{ - 1}}\)    A1

consider \(f(a * b) = {(a * b)^{ - 1}}\)     M1

\( = {b^{ - 1}} * {a^{ - 1}}\)    A1

\( = {a^{ - 1}} * {b^{ - 1}}\) since \(G\) is Abelian     R1

hence \(f\) is an isomorphism     AG

[7 marks]

A surprising number of candidates wasted time and unrewarded effort showing that the mapping \(f\), stated to be a bijection in the question, actually was a bijection. Many candidates failed to get full marks by not properly using the fact that the group was stated to be Abelian. There were also candidates who drew the graph of \(y = \frac{1}{x}\) or otherwise assumed that the inverse of \(x\) was its reciprocal - this is unacceptable in the context of an abstract group question.