= Solution
For $|G|=p^rm$ 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 $n_p\equiv1\pmod p$. 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 \b[every <group> of order $p^2$ is abelian].
Now consider $|G|=pq^a$ with $0\leq a\leq2$. 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 $a\leq1$, $n_p$ 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 \b[all <groups> of the required orders are soluble], by elementary Sylow arguments alone.
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