Find a \(2\)-Sylow subgroup of \(S_5\).
Find a \(5\)-Sylow subgroup of \(S_{15}\).
For prime \(p\), how many \(p\)-Sylow subgroups does \(S_p\) have?
For prime \(p\), what is the order of the normalizer of a \(p\)-Sylow subgroup \(P\) in \(S_p\)?
Solution
1. A Sylow \(2\)-subgroup of \(S_5\) is\[ P_2=\langle (1\,2\,3\,4),(1\,3)\rangle. \]
Let \(r=(1\,2\,3\,4)\) and \(s=(1\,3)\). Then \(r^4=s^2=1\) and \[ srs=r^{-1}, \] so \(\langle r,s\rangle\cong D_8\) has order \(8\). Since \[ |S_5|=120=2^3\cdot 3\cdot 5, \] a subgroup of order \(2^3\) is Sylow.
2. A Sylow \(5\)-subgroup of \(S_{15}\) is\[ P_5=\langle (1\,2\,3\,4\,5), (6\,7\,8\,9\,10), (11\,12\,13\,14\,15) \rangle\cong C_5^3. \]
The three displayed \(5\)-cycles are disjoint, hence commute and generate a subgroup of order \(5^3\). Also \[ v_5(15!)=\left\lfloor\frac{15}{5}\right\rfloor+ \left\lfloor\frac{15}{25}\right\rfloor=3. \] Thus the largest power of \(5\) dividing \(|S_{15}|\) is \(5^3\), so \(P_5\) is Sylow.
3. For every prime \(p\), the number of Sylow \(p\)-subgroups of \(S_p\) is\[ n_p=(p-2)!. \]
Since \(v_p(p!)=1\), a Sylow \(p\)-subgroup has order \(p\) and is generated by a \(p\)-cycle. The number of \(p\)-cycles in \(S_p\) is \((p-1)!\). Each subgroup of order \(p\) has exactly \(p-1\) nonidentity elements, and every nonidentity element is a \(p\)-cycle. Distinct subgroups of order \(p\) intersect trivially, so \[ n_p=\frac{(p-1)!}{p-1}=(p-2)!. \] For \(p=2\), this gives \(0!=1\).
4. If \(P\) is a Sylow \(p\)-subgroup of \(S_p\), then\[ |N_{S_p}(P)|=p(p-1). \]
The Sylow \(p\)-subgroups form a single conjugacy class, and the stabilizer of \(P\) under conjugation is its normalizer. Hence orbit-stabilizer gives \[ [S_p:N_{S_p}(P)]=n_p=(p-2)!. \] Therefore \[ |N_{S_p}(P)|= \frac{p!}{(p-2)!}=p(p-1). \]