(i)     Show that $gh{g^{ - 1}}$ has order 2 for all $g \in G$.

(ii)     Deduce that gh = hg for all $g \in G$.

(i)     consider ${(gh{g^{ - 1}})^2}$     M1

$= gh{g^{ - 1}}gh{g^{ - 1}} = g{h^2}{g^{ - 1}} = g{g^{ - 1}} = e$     A1

$gh{g^{ - 1}}$ cannot be order 1 (= e) since h is order 2     R1

so $gh{g^{ - 1}}$ has order 2     AG

(ii)     but h is the unique element of order 2     R1

hence $gh{g^{ - 1}} = h \Rightarrow gh = hg$     A1AG

[5 marks]

This question was by far the problem to be found most challenging by the candidates. Many were able to show that $gh{g^{ - 1}}$ had order one or two although hardly any candidates also showed that the order was not one thus losing a mark. Part a (ii) was answered correctly by a few candidates who noticed the equality of h and $gh{g^{ - 1}}$. However, many candidates went into algebraic manipulations that led them nowhere and did not justify any marks. Part (b) (i) was well answered by a small number of students who appreciated the nature of the identity and element h thus forcing the other two elements to have order four. However, (ii) was only occasionally answered correctly and even in these cases not systematically. It is possible that candidates lacked time to fully explore the problem. A small number of candidates “guessed” the correct answer.