Past exam of the algebra course of the University of Oxford / 2007 / B / 5

For $|G|=p^{r}m$ with $p\nmid m$, Sylow's theorem gives subgroups of order $p^r$, says that every $p$-subgroup lies in a conjugate of one of them, and that all such Sylow subgroups are conjugate. Their number $n_p$ divides $m$ and satisfies $n_p\equiv 1(\mathrm{mod}p)$. A group is soluble when its derived series, obtained by repeatedly taking the commutator subgroup, reaches the identity. Equivalently it has a subnormal series with abelian factors.

For a group of order $p^2$, the class equation shows its center is nontrivial: noncentral conjugacy classes have sizes divisible by $p$. If the center had order $p$, the quotient by the center would be cyclic; a group with cyclic central quotient is abelian, since representatives are powers of one element times central elements. This would make the center the whole group. Thus every group of order $p^2$ is abelian.

Now consider $|G|=p{q}^a$ with $0\le a\le 2$. Cases with a normal Sylow subgroup are soluble: that subgroup has prime or prime-square order and is abelian, and its quotient also has such order and is abelian. This gives a subnormal series with abelian factors.

If $p<q$, the Sylow number $n_q$ divides $p$ and is $1$ modulo $q$, so is one. If $p>q$ and $a\le 1$, $n_p$ divides $q^a<p$, so again is one. For $p>q$ and $a=2$, a nontrivial $n_p$ can only be $q^2$, since $q<p$. The congruence would force $p\mid (q-1)(q+1)$, hence $p\mid q+1$. Since $p>q$, this means $p=q+1$, and primality forces $(p,q)=(3,2)$.

In the remaining order-twelve case, either a Sylow subgroup is already normal, or there are four Sylow subgroups of order three. Those have pairwise trivial intersection and contribute eight distinct elements of order three. Only four elements remain, including the identity. Every Sylow subgroup of order four lies entirely among those four elements, so it is unique and normal. It and its quotient are abelian. Therefore all groups of the required orders are soluble, by elementary Sylow arguments alone.