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.