The group $\{ G,{\text{ }} * \}$ is defined on the set $G$ with binary operation $*$. $H$ is a subset of $G$ defined by $H = \{ x:{\text{ }}x \in G,{\text{ }}a * x * {a^{ - 1}} = x{\text{ for all }}a \in G\}$. Prove that $\{ H,{\text{ }} * \}$ is a subgroup of $\{ G,{\text{ }} * \}$.

associativity: This follows from associativity in $\{ G,{\text{ }} * \}$     R1

the identity $e \in H$ since $a * e * {a^{ - 1}} = a * {a^{ - 1}} = e$ (for all $a \in G$)     R1

Note:     Condone the use of the commutativity of e if that is involved in an alternative simplification of the LHS.

closure: Let $x,{\text{ }}y \in H$ so that $a * x * {a^{ - 1}} = x$ and $a * y * {a^{ - 1}} = y$ for all $a \in G$     (M1)

multiplying, $x * y = a * x * {a^{ - 1}} * a * y * {a^{ - 1}}$ (for all $a \in G$)     A1

$= a * x * y * {a^{ - 1}}$    A1

therefore $x * y \in H$ (proving closure)     R1

inverse: Let $x \in H$ so that $a * x * {a^{ - 1}} = x$ (for all $a \in G$)

${x^{ - 1}} = {(a * x * {a^{ - 1}})^{ - 1}}$    M1

$= a * {x^{ - 1}} * {a^{ - 1}}$    A1

therefore ${x^{ - 1}} \in H$     R1

hence $\{ H,{\text{ }} * \}$ is a subgroup of $\{ G,{\text{ }} * \}$     AG

Note:     Accuracy marks cannot be awarded if commutativity is assumed for general elements of $G$.

[9 marks]

This is an abstract question, clearly defined on a subset. Far too many candidates almost immediately deduced, erroneously, that the full group was Abelian. Almost no marks were then available.