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.