- Let \(G\) be a finite group of order \(p^e v\), where \(e\) and \(v\) are positive integers, \(p\) is prime, \(p>v\), and \(p\nmid v\). Show that \(G\) has a normal Sylow p-subgroup.
Solution
Let \(n_p\) denote the number of Sylow \(p\)-subgroups of \(G\).
1. Sylow’s theorem gives\[ n_p\mid v, \qquad n_p\equiv1\pmod p. \]
Since \[ |G|=p^e v \] with \(p\nmid v\), a Sylow \(p\)-subgroup has order \(p^e\). The third Sylow theorem says that the number of such subgroups divides the complementary factor \(v\) and is congruent to \(1\) modulo \(p\).
2. One has \(n_p=1\).
Because \(n_p\mid v\), we have \[ 1\le n_p\le v<p. \] The only positive integer strictly less than \(p\) that is congruent to \(1\) modulo \(p\) is \(1\). Thus \(n_p=1\).
3. Therefore the Sylow \(p\)-subgroup is normal.
All Sylow \(p\)-subgroups are conjugate. Since there is exactly one, it is fixed by conjugation by every element of \(G\), hence is normal.