(i)      Prove that \(G\) is cyclic and state two of its generators.

(ii)     Let \(H\) be the subgroup generated by \({a^4}\). Construct a Cayley table for \(H\).

State, with a reason, whether or not it is necessary that a group is cyclic given that all its proper subgroups are cyclic.

(i)   the order of \(a\) is a divisor of the order of \(G\)     (M1)

since the order of \(G\) is 12, the order of \(a\) must be 1, 2, 3, 4, 6 or 12     A1

the order cannot be 1, 2, 3 or 6, since \({a^6} \ne e\)     R1

the order cannot be 4, since \({a^4} \ne e\)     R1

so the order of \(a\) must be 12

therefore, \(a\) is a generator of \(G\), which must therefore be cyclic     R1

another generator is eg \({a^{ - 1}},{\text{ }}{a^5},{\text{ }} \ldots \)     A1

[6 marks]

(ii)     \(H = \{ e,{\text{ }}{a^4},{\text{ }}{a^8}\} \)     (A1)

     M1A1

[3 marks]

no     A1

eg the group of symmetries of a triangle \({S_3}\) is not cyclic but all its (proper) subgroups are cyclic

eg the Klein four-group is not cyclic but all its (proper) subgroups are cyclic     R1

[2 marks]

In part (a), many candidates could not provide a logical sequence of steps to show that \(G\) is cyclic. In particular, although they correctly quoted Lagrange’s theorem, they did not always consider all the orders of a, i.e., all the factors of 12, omitting in particular 1 as a factor. Some candidates did not state the second generator, in particular \({a^{ - 1}}\). Very few candidates were successful in finding the required subgroup, although they were obviously familiar with setting up a Cayley table.