Sylow theory is the main tool for extracting subgroup information from the prime factorization of a finite group order. It turns divisibility into existence, conjugacy, and counting constraints.
p-groups and Sylow subgroups
Definition. A finite group is a \(p\)-group if its order is a power of a prime \(p\). If \(|G|=p^am\) with \(p\nmid m\), a Sylow \(p\)-subgroup is a subgroup of order \(p^a\).
Lemma. If \(G\) is a finite \(p\)-group, then \(Z(G)\ne\{e\}\). Consequently every group of order \(p^2\) is abelian.
Proof. The nontrivial center follows from the class equation. If \(|G|=p^2\), then \(|Z(G)|\) is \(p\) or \(p^2\). If it were \(p\), then \(G/Z(G)\) would be cyclic, which forces \(G\) abelian, contradicting \(|Z(G)|=p\).
The Sylow theorems
Theorem. Let \(|G|=p^am\) with \(p\nmid m\). Then: Sylow \(p\)-subgroups exist; any two are conjugate; the number \(n_p\) of Sylow \(p\)-subgroups satisfies \[ n_p\equiv1\pmod p, \qquad n_p\mid m. \]
Remark. The congruence and divisibility conditions are often more useful than the existence statement: they frequently force \(n_p=1\), which means the Sylow subgroup is normal.
Normal Sylow subgroups
Proposition. A Sylow \(p\)-subgroup \(P\) is normal in \(G\) iff it is the unique Sylow \(p\)-subgroup.
Proof. If \(P\trianglelefteq G\), every conjugate \(gPg^{-1}\) equals \(P\), and Sylow conjugacy says all Sylow \(p\)-subgroups are conjugates of \(P\). Conversely, uniqueness makes \(gPg^{-1}=P\) for every \(g\).
Example. A group of order \(15=3\cdot5\) has \(n_5\equiv1\pmod5\) and \(n_5\mid3\), so \(n_5=1\). It also has \(n_3\equiv1\pmod3\) and \(n_3\mid5\), so \(n_3=1\) or \(10\); the divisibility condition rules out \(10\), hence \(n_3=1\). Both Sylow subgroups are normal.
Groups of order pq
Theorem. Let \(p<q\) be primes. If \(p\nmid(q-1)\), then every group of order \(pq\) is cyclic. If \(p\mid(q-1)\), there may also be a nonabelian semidirect product.
Example. Groups of order \(21\) need not be cyclic because \(3\mid(7-1)\). A nontrivial homomorphism \(C_3\to\operatorname{Aut}(C_7)\cong C_6\) produces a nonabelian semidirect product \(C_7\rtimes C_3\).
Non-simplicity tests
Proposition. If the Sylow counting conditions force \(n_p=1\) for some prime divisor \(p\) of \(|G|\), then \(G\) is not simple unless the Sylow subgroup is trivial or all of \(G\).
Calculation. If \(|G|=45=3^2\cdot5\), then \(n_5\equiv1\pmod5\) and \(n_5\mid9\). Hence \(n_5=1\), so the subgroup of order \(5\) is normal and \(G\) is not simple.
How to use the theorem without overusing it
Remark. Sylow theory gives subgroup counts but not automatically the whole group structure. Once normal Sylow subgroups are known, one still needs to decide how the subgroups combine, usually through direct or semidirect products.