State the Sylow Theorems. Include information about the existence, order, and number \(n_p\) of Sylow \(p\)-subgroups of a finite group, and mention additional divisibility and congruence properties of \(n_p\).
Solution
Let \(G\) be finite, let \(p\) be prime, and write \(|G|=p^a m\) with \(a\geq0\) and \(p\nmid m\).
1. A Sylow \(p\)-subgroup is a subgroup of order \(p^a\).Such subgroups exist. Every subgroup of \(G\) whose orderis a power of \(p\) is contained in a Sylow \(p\)-subgroup.Any two Sylow \(p\)-subgroups are conjugate by an elementof \(G\) [@DF04, sec. 4.5].
2. The number \(n_p\) of Sylow \(p\)-subgroups satisfies\[ n_p=[G:N_G(P)],\qquad n_p\mid m,\qquad n_p\equiv1\pmod p \]for any Sylow \(p\)-subgroup \(P\).
Conjugation is transitive on the Sylow subgroups by step 1, and the stabilizer of \(P\) is its normalizer \(N_G(P)=\{g\in G:gPg^{-1}=P\}\). Orbit-stabilizer gives the index formula [@DF04]. Since \(P\leq N_G(P)\), \[ m=[G:P]=[G:N_G(P)]\,[N_G(P):P], \] so \(n_p\mid m\).
For the congruence, restrict the conjugation action to \(P\). Every orbit has power-of-\(p\) size by orbit-stabilizer; hence all non-singleton orbits have size divisible by \(p\). The subgroup \(P\) itself is a fixed point. If another Sylow subgroup \(Q\) is fixed, then \(P\) normalizes \(Q\). Consequently \(PQ\) is a subgroup, of order \(|P||Q|/|P\cap Q|\), a power of \(p\). Its order divides \(|G|\) and hence is at most \(p^a\). But it contains both order-\(p^a\) groups \(P,Q\), forcing \(P=PQ=Q\). Thus there is exactly one fixed point, and counting orbits gives \(n_p\equiv1\pmod p\).
3. A Sylow \(p\)-subgroup \(P\) is normal exactly when \(n_p=1\).
If \(P\) is normal, its conjugacy orbit has one member; conjugacy in step 1 then implies that it is the only Sylow subgroup. Conversely, if it is unique, each conjugate has the same order and must equal \(P\), which is normality. For \(a=0\), the unique Sylow \(p\)-subgroup is the trivial group.