Let \((H,{\text{ }} * {\text{)}}\) be a subgroup of the group \((G,{\text{ }} * {\text{)}}\).

Consider the relation \(R\) defined in \(G\) by \(xRy\) if and only if \({y^{ - 1}} * x \in H\).

(a)     Show that \(R\) is an equivalence relation on \(G\).

(b)     Determine the equivalence class containing the identity element.

(a)     \(R\) is reflexive as \({x^{ - 1}} * x = e \in H \Rightarrow xRx\) for any \(x \in G\)     A1

if \(xRy\) then \({y^{ - 1}} * x = h \in H\)

but \(h \in H \Rightarrow {h^{ - 1}} \in H\), ie, \(\underbrace {{{({y^{ - 1}} * x)}^{ - 1}}}_{{x^{ - 1}}{\text{*}}y} \in H\)     M1

therefore \(yRx\)

\(R\) is symmetric     A1

if \(xRy\) then \({y^{ - 1}} * x = h \in H\) and if \(yRz\) then \({z^{ - 1}} * y = k \in H\)     M1

\(k * h \in H\), ie, \(\underbrace {({z^{ - 1}} * y) * ({y^{ - 1}} * x)}_{{z^{ - 1}} * x} \in H\)     A1

therefore \(xRz\)

\(R\) is transitive     A1

so \(R\) is an equivalence relation on \(G\)     AG

[6 marks]

(b)     \(xRe \Leftrightarrow {e^{ - 1}} * x \in H\)     M1

\(x \in H\)     A1

\([e] = H\)     A1     N0

[3 marks]

Part (a) was fairly well answered by many candidates. They knew how to apply the equivalence relations axioms in this particular example. Part (b) however proved to be very challenging and hardly any correct answers were seen.