Let \(G\) be a group of order \(p^k m\), where \(p\) is a prime not dividing \(m\).
What is a Sylow \(p\)-subgroup of \(G\)?
State Sylow’s theorem, including information about the number of Sylow \(p\)-subgroups of \(G\) and about relationships between them.
Show that there are no simple groups of order 132.
Solution
1. A Sylow \(p\)-subgroup is a subgroup of order \(p^k\).Sylow’s theorems assert that such subgroups exist, that every\(p\)-subgroup is contained in one, and that all Sylow \(p\)-subgroupsare conjugate. Their number \(n_p\) satisfies\[ n_p\mid m,\qquad n_p\equiv1\pmod p. \]These are the Sylow theorems in their finite-group form [@DF04].In particular, a unique Sylow subgroup is normal, because conjugationpreserves its order.
2. A group of order \(132=2^2\cdot3\cdot11\) has a nontrivialproper normal subgroup.
The number \(n_{11}\) divides \(12\) and is \(1\) modulo \(11\), so \(n_{11}=1\) or \(12\). In the first case the unique subgroup of order \(11\) is the required normal subgroup.
Suppose \(n_{11}=12\). Two different subgroups of order \(11\) intersect only in the identity, since a nonidentity element generates each subgroup of prime order containing it. They therefore account for \(12\cdot10=120\) distinct elements of order \(11\). Of the \(131\) nonidentity elements in \(G\), only \(11\) remain. Different subgroups of order \(3\) likewise have disjoint nonidentity parts, and each contributes two elements of order \(3\). Hence \(2n_3\leq11\), giving \(n_3\leq5\).
By Sylow’s theorem, \(n_3\) divides \(44\) and is \(1\) modulo \(3\). The divisors of \(44\) are \(1,2,4,11,22,44\); the ones congruent to \(1\) modulo \(3\) are \(1,4,22\). The bound just obtained therefore leaves \(n_3=1\) or \(4\).
If \(n_3=1\), the unique subgroup of order \(3\) is normal and is nontrivial and proper. If \(n_3=4\), conjugation on these four subgroups gives a homomorphism \(\varphi:G\to S_4\). It is not injective, since \(|G|=132>24=|S_4|\), so \(\ker\varphi\ne\{1\}\). Its image is not trivial: Sylow conjugacy makes the action transitive on a set of four elements, whereas the trivial action has singleton orbits. Thus \(\ker\varphi\ne G\) as well. Being the kernel of a homomorphism, it is a normal subgroup. Every case has produced a nontrivial proper normal subgroup, so no group of order \(132\) is simple.