State/prove the Sylow theorems.
Solution
Let \(G\) be finite and write \[ |G|=p^a m, \qquad p\nmid m. \] A subgroup of order \(p^a\) is called a Sylow \(p\)-subgroup.
1. First Sylow theorem. A Sylow \(p\)-subgroup of \(G\) exists.
We argue by induction on \(|G|\). The result is immediate for \(|G|=1\). Assume \(p\mid |G|\).
If \(p\mid |Z(G)|\), Cauchy’s theorem gives a central subgroup \(C\le Z(G)\) of order \(p\). Then \[ |G/C|=p^{a-1}m. \] By induction, \(G/C\) has a subgroup \(\overline P\) of order \(p^{a-1}\). Its inverse image \(P\) in \(G\) has order \(p^a\).
Now suppose \(p\nmid |Z(G)|\). The class equation is \[ |G|=|Z(G)|+\sum_i [G:C_G(x_i)], \] where the \(x_i\) represent the noncentral conjugacy classes. If every index \([G:C_G(x_i)]\) were divisible by \(p\), then the right side would be congruent to \(|Z(G)|\not\equiv0\pmod p\), contradicting \(p\mid|G|\). Thus for some noncentral \(x\), \[ p\nmid [G:C_G(x)]. \] Hence the full \(p\)-part \(p^a\) of \(|G|\) divides \(|C_G(x)|\). Since \(x\) is noncentral, \(C_G(x)<G\), so induction applied to \(C_G(x)\) gives a subgroup of order \(p^a\). This is a Sylow \(p\)-subgroup of \(G\).
2. Second Sylow theorem. If \(P\) is a Sylow \(p\)-subgroup and \(H\le G\) is any \(p\)-subgroup, then \(H\) is contained in a conjugate of \(P\). In particular, all Sylow \(p\)-subgroups are conjugate.
Let \(H\) act by left multiplication on the set \(G/P\) of left cosets. The number of cosets is \[ [G:P]=m, \] which is not divisible by \(p\). Every \(H\)-orbit has size a power of \(p\). If every orbit had size divisible by \(p\), then \(p\) would divide \(|G/P|=m\), impossible. Thus there is a fixed coset \(gP\).
The equality \[ hgP=gP\qquad(h\in H) \] is equivalent to \[ g^{-1}hg\in P\qquad(h\in H). \] Hence \[ g^{-1}Hg\le P, \] or equivalently \(H\le gPg^{-1}\).
If \(H\) is itself Sylow, then \(|H|=|P|\), so containment forces equality. Thus every Sylow \(p\)-subgroup is conjugate to \(P\).
3. Third Sylow theorem. If \(n_p\) denotes the number of Sylow \(p\)-subgroups, then\[ n_p\mid m \qquad\text{and}\qquad n_p\equiv1\pmod p. \]
By 2, the Sylow \(p\)-subgroups form one conjugacy orbit. The stabilizer of \(P\) under conjugation is its normalizer \(N_G(P)\), so orbit-stabilizer gives \[ n_p=[G:N_G(P)]. \] Since \(P\le N_G(P)\), \[ [G:P]=[G:N_G(P)]\,[N_G(P):P]=n_p[N_G(P):P]. \] Thus \(n_p\mid [G:P]=m\).
For the congruence, let \(P\) act by conjugation on the set \(\operatorname{Syl}_p(G)\). Every orbit has size a power of \(p\). We claim that \(P\) is the unique fixed point. Certainly \(P\) fixes itself. If another Sylow subgroup \(Q\) is fixed, then \(P\le N_G(Q)\). Since \(Q\trianglelefteq N_G(Q)\), the product \(PQ\) is a subgroup of \(N_G(Q)\); it is a \(p\)-subgroup of \(G\). Maximality of the Sylow subgroup \(Q\) forces \(PQ=Q\), hence \(P\le Q\). Since \(|P|=|Q|\), we get \(P=Q\).
Therefore all orbits except the singleton \(\{P\}\) have cardinality divisible by \(p\), so \[ n_p\equiv1\pmod p. \]
These are the three Sylow theorems.