Let \(G\) be a group of order \[ 231=3\cdot7\cdot11. \] Show that \(G\) has a normal Sylow \(7\)-subgroup and that its Sylow \(11\)-subgroup is central.
Solution
1. The Sylow \(7\)-subgroup is normal.
Let \(n_7\) be the number of Sylow \(7\)-subgroups. Sylow gives \[ n_7\equiv1\pmod7, \qquad n_7\mid 33. \] The divisors of \(33\) are \(1,3,11,33\), and only \(1\) is congruent to \(1\) modulo \(7\). Hence \[ n_7=1. \] So the Sylow \(7\)-subgroup is unique and therefore normal.
2. The Sylow \(11\)-subgroup is unique.
Similarly, \[ n_{11}\equiv1\pmod{11}, \qquad n_{11}\mid21. \] Among \(1,3,7,21\), only \(1\) is congruent to \(1\) modulo \(11\). Thus \(n_{11}=1\). Let \(Q\) denote this unique Sylow \(11\)-subgroup; then \(Q\trianglelefteq G\) and \(Q\cong C_{11}\).
3. The subgroup \(Q\) is central.
Conjugation gives a homomorphism \[ G\longrightarrow\Aut(Q)\cong C_{10}. \] Because \(Q\) is abelian, \(Q\) lies in the kernel, so the image has order dividing \[ |G/Q|=21. \] It also has order dividing \(|\Aut(Q)|=10\). Since \[ \gcd(21,10)=1, \] the image is trivial. Thus every element of \(G\) centralizes every element of \(Q\), so \[ Q\subseteq Z(G). \]