(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.