Count the number of \(p\)-Sylow subgroups of \(S_p\).
Solution
For every prime \(p\), the number is \(\boxed{(p-2)!}\).
1. Each Sylow \(p\)-subgroup is cyclic of order \(p\),and its nonidentity elements are \(p\)-cycles.
The order of \(S_p\) is \(p!=p(p-1)!\), and \(p\) does not divide \((p-1)!\). Sylow subgroups therefore have order \(p\) [@DF04]. A nonidentity element generates any group of prime order by Lagrange’s theorem.
The order of a permutation is the least common multiple of its disjoint-cycle lengths [@DF04]. For order \(p\), every cycle length is \(1\) or \(p\), and at least one is \(p\). On exactly \(p\) letters, the permutation must thus be one \(p\)-cycle. Conversely, a \(p\)-cycle generates a subgroup of order \(p\), which is Sylow.
2. Dividing the number of \(p\)-cycles by \(p-1\)counts the Sylow subgroups exactly once.
There are \((p-1)!\) distinct \(p\)-cycles: write each uniquely as \((1,a_2,\ldots,a_p)\), with the other \(p-1\) letters in arbitrary order. Each cyclic subgroup of order \(p\) has \(p-1\) nonidentity elements, all generators. Distinct such subgroups cannot share one, since that element would generate both. By step 1 these disjoint generator sets partition the \(p\)-cycles. Thus the number is \[ \frac{(p-1)!}{p-1}=(p-2)!. \] For \(p=2\) this is \(0!=1\), the subgroup \(S_2\) itself.