Show that any group of order \(p^2q\) (for primes \(p,q\)) is solvable.
Solution
Let \(|G|=p^2q\) with \(p,q\) prime.
1. If \(p=q\), then \(G\) is solvable.
In this case \(|G|=p^3\), so \(G\) is a finite \(p\)-group. Every finite \(p\)-group is nilpotent, hence solvable.
Assume henceforth that \(p\ne q\).
2. At least one Sylow subgroup of \(G\) is normal.
Let \(n_p\) and \(n_q\) denote the numbers of Sylow \(p\)- and \(q\)-subgroups. Sylow’s theorems give \[ n_p\mid q,\qquad n_p\equiv1\pmod p, \] and \[ n_q\mid p^2,\qquad n_q\equiv1\pmod q. \]
If \(p>q\), then \(n_p\) is either \(1\) or \(q\), but \(q<p\) prevents \(q\equiv1\pmod p\). Thus \(n_p=1\).
Suppose \(p<q\). If \(n_q=1\), we are done. Otherwise \(n_q\) is \(p\) or \(p^2\). Since \(p<q\), the congruence \(n_q\equiv1\pmod q\) excludes \(n_q=p\), so \(n_q=p^2\). Hence \[ q\mid p^2-1=(p-1)(p+1). \] Because \(q>p\), the prime \(q\) cannot divide \(p-1\), so \(q\mid p+1\). Thus \(q=p+1\). One of \(p\) and \(p+1\) is even, so the only primes with \(q=p+1\) are \(p=2\), \(q=3\).
It remains to consider \(|G|=12\). If \(n_3=4\), then the four Sylow \(3\)-subgroups have pairwise trivial intersection and contribute \[ 4(3-1)=8 \] nonidentity elements. Only three nonidentity elements remain. Any Sylow \(2\)-subgroup has order \(4\) and hence already contains three nonidentity elements, so every Sylow \(2\)-subgroup must consist of the identity together with those same three remaining elements. Therefore the Sylow \(2\)-subgroup is unique and normal. Thus in all cases at least one Sylow subgroup is normal.
3. A normal Sylow subgroup \(N\) of \(G\) is solvable, and so is \(G/N\).
The possible orders of \(N\) are \(p^2\) or \(q\). A group of prime order is cyclic, and every group of order \(p^2\) is abelian. Hence \(N\) is abelian and therefore solvable.
The quotient has the complementary order: if \(|N|=p^2\), then \(|G/N|=q\); if \(|N|=q\), then \(|G/N|=p^2\). Thus \(G/N\) is likewise abelian and solvable.
4. Therefore \(G\) is solvable.
A group with a solvable normal subgroup and solvable quotient is solvable. Apply this extension criterion to the normal Sylow subgroup supplied by 2 and use 3.