Prove there is no simple group of order 132.
Solution
Suppose for contradiction that \(G\) is simple and \[ |G|=132=2^2\cdot3\cdot11. \]
1. The number \(n_{11}\) of Sylow \(11\)-subgroups is \(12\).
Sylow gives \[ n_{11}\mid12,\qquad n_{11}\equiv1\pmod{11}. \] Thus \(n_{11}=1\) or \(12\). Simplicity excludes \(n_{11}=1\), so \(n_{11}=12\).
2. These Sylow \(11\)-subgroups contribute \(120\) distinct nonidentity elements.
Each has \(10\) nonidentity elements. Two distinct subgroups of order \(11\) intersect trivially, because their intersection has order dividing the prime \(11\) and cannot be all of either subgroup. Hence the \(12\) subgroups contribute \[ 12(11-1)=120 \] distinct nonidentity elements.
3. The number \(n_3\) of Sylow \(3\)-subgroups must be \(4\).
Sylow gives \[ n_3\mid44,\qquad n_3\equiv1\pmod3. \] Thus \(n_3\in\{1,4,22\}\). Simplicity excludes \(1\). If \(n_3=22\), the distinct order-\(3\) subgroups would contribute \[ 22(3-1)=44 \] nonidentity elements. But after the \(120\) nonidentity elements of order \(11\) from 2, only \(11\) nonidentity elements remain. Contradiction. Hence \(n_3=4\).
4. Conjugation on the four Sylow \(3\)-subgroups gives a nontrivial homomorphism\[ \varphi:G\to S_4. \]
The conjugation action of \(G\) on its Sylow \(3\)-subgroups is transitive by Sylow conjugacy. Since there are four of them, it defines \(\varphi:G\to S_4\). The action is not trivial because a trivial action would fix every Sylow \(3\)-subgroup, making each one normal, contrary to simplicity.
5. Simplicity would force \(\varphi\) to be injective, which is impossible.
The kernel is a normal subgroup of the simple group \(G\). Since \(\varphi\) is nontrivial, the kernel is not all of \(G\), so it must be trivial. Hence \(G\) would embed in \(S_4\). But \[ |G|=132>24=|S_4|, \] which is impossible.
6. Therefore no group of order \(132\) is simple.
The assumption of simplicity led to the contradiction in 5.