Let \(p,q\) be distinct primes, and let \(k\) denote the smallest positive integer such that \(p\) divides \(q^k - 1\). Show that no group of order \(pq^k\) is simple.
Let \(p,q\) be distinct primes, and let \(k\) denote the smallest positive integer such that \(p\mid q^k-1\). Show that no group of order \(pq^k\) is simple.
Solution
Let \(G\) have order \(pq^k\).
Let \(n_p\) be the number of Sylow \(p\)-subgroups. Sylow gives \[ n_p\mid q^k, \qquad n_p\equiv1\pmod p. \] Hence \(n_p=q^j\) for some \(0\le j\le k\). If \(j=0\), then \(n_p=1\), so the Sylow \(p\)-subgroup is normal and \(G\) is not simple.
Suppose therefore that \(j>0\). Since \(q^j\equiv1\pmod p\), the minimality of \(k\) forces \(j\ge k\). Thus \(j=k\) and \[ n_p=q^k. \] Distinct Sylow \(p\)-subgroups have trivial intersection because they have prime order. Therefore they contribute \[ q^k(p-1) \] nonidentity elements. Since \(|G|=pq^k\), exactly \[ pq^k-q^k(p-1)=q^k \] elements remain.
Any Sylow \(q\)-subgroup has order \(q^k\) and contains no nonidentity element of order \(p\), so it is contained in this remaining set of \(q^k\) elements. Hence there is only one Sylow \(q\)-subgroup. It is normal.
In all cases \(G\) has a nontrivial proper normal Sylow subgroup, so \(G\) is not simple.