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If $G$ is a finite group, and $P$ is a Sylow-$p$ subgroup of $G$, then the number of Sylow-$p$ subgroups in $G$ is at most $|G|/|P|$.

In the Symmetric group $S_n$, the bound is attained only for certain Sylow subgroups.

In $S_4$, the number of Sylow-$2$ subgroups is $|S_4|/2^3$.

In $S_3$, the number of Sylow-$3$ subgroups is less than $|S_3|/3$.

The question I want to pose here is then the natural one:

Question: What are the conditions on $n$ and $p$ which are necessary and/or sufficient to say that the number of Sylow-$p$ subgroups in $S_n$ is $|S_n|/|P|$, where $P$ is a Sylow-$p$ subgroup of $S_n$.

Groups
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2 Answers2

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The condition is equivalent to a Sylow $p$-subgroup being self-normalizing in $S_n$, and the answer is very simple (simple to state, anyway): $p=2$ is both necessary and sufficient for the condition to hold for all $n$.

It is well-known that a Sylow $p$-subgroup $P$ of $S_n$ is a direct product of iterated wreath products $P_k = C_p \wr C_p \wr \cdots \wr C_p$ (with $k$ wreath factors), where the direct factors $P_k$ have disjoint supports of size $p^k$.

Now $P_k$ is normalized in $S_{p^k}$ by $P_k = C_p \wr C_p \wr \cdots \wr P_k \wr F_{p(p-1)}$, where the final wreath factor $F_{p(p-1)}$ is a Frobenius group of order $p(p-1)$. So for $p$ odd, the Sylow $p$-subgroup $P$ is strictly contained in its normalizer.

When $p=2$, we have $P = P_{k_1} \times \cdots \times P_{k_j}$, where $n = 2^{k_1} + \cdots + 2^{k_j}$ with all $k_i$ distinct, so the orbits of $P$ all have different lengths. Hence the normalizer of $P$ in $S_n$ fixes all of the orbits, and so it suffices to prove that $P_k$ is self-normalizing in $S_{2^k}$.

We could try and prove this by induction on $k$. Now $P_k = C_2 \wr P_{k-1}$ with base-group $N$, which is elementary abelian of order $2^{2^{k-1}}$ and is generated by the transpositions in $P_k$. So it must be normalized by the normaliser of $P_k$ in $S_n$. Now the normalizer of $N$ must permute these transpositions so it is equal to $C_2 \wr S_{2^{k-1}}$, and now we want the normaliser of $C_2 \wr P_{k-1}$ in $C_2 \wr S_{2^{k-1}}$, and we can use induction.

Derek Holt
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Maybe it's worth recalling the formula for the number $n_p(S_n)$ of Sylow $p$-subgroups of $S_n,$ for all $n$ and all primes $p.$ Let $n=n_0+\cdots n_kp^k$ with $0\le n_i<p$ be the base $p$ expansion of $n;$ then $$n_p(S_n)=\frac{(n!)_{p^\prime}}{n_0!\cdots n_k!\cdot (p-1)^{n_1+2n_2+\cdots+kn_k}},$$ where $(n!)_{p^\prime}$ means the largest factor of $n!$ that is prime to $p.$

I noticed this while considering this question. The formula will be in a textbook somewhere, but I wasn't aware of it and it's reasonably memorable. It's nice to see immediately that $n_p(S_n)=(n!)_{p^\prime}$ iff $p=2$ or $n=1,$ as expected.

The formula follows from the theory explained in Derek Holt's answer. In fact we just need to know that the center of a Sylow $p$-subgroup of $S_n$ is elementary abelian and contains $(p-1)^{n_1+\cdots+n_k}$ elements in the conjugacy class $\mathcal C$ defined by cycle type $1^{n_0}p^{[n/p]}.$ Hence the number of pairs $(P,\pi)$ with $P$ a Sylow $p$-subgroup of $G$ and $\pi\in\mathcal C$ centralized by $P$ is $n_p(S_n)(p-1)^{n_1+\cdots+n_k}.$ Counting this number the other way gives $$n_p(S_n)(p-1)^{n_1+\cdots+n_k}=|\mathcal C|n_p(C_G(\pi)).$$ However $C_G(\pi)/C_p^{[n/p]}\cong S_{[n/p]}\times S_{n_0},$ so $n_p(C_G(\pi))=n_p(S_{[n/p]}).$ Since $|\mathcal C|=n!/(n_0![n/p]!p^{[n/p]})$ we get $$n_p(S_n)=\frac{n!}{n_0![n/p]!p^{[n/p]}(p-1)^{n_1+\cdots+n_k}}n_p(S_{[n/p]}),$$ and the formula now follows by induction.

Tom WIlde
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