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 $S_n$.


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.

  • $\begingroup$ Interesting. I would try to prove this. $\endgroup$ – Groups Jul 28 '15 at 9:48
  • $\begingroup$ Surely this needs justification (or at least a link to the relevant paper) $\endgroup$ – Prahlad Vaidyanathan Jul 28 '15 at 9:48
  • $\begingroup$ Yes it needs justification doesn't it! I will leave that as an exercise, but the Sylow $p$-subgroups of $S_n$ are easily described as iterated wreath products, and the proof reduces easily to the case when $n$ is a power of $p$. $\endgroup$ – Derek Holt Jul 28 '15 at 9:50
  • $\begingroup$ @PrahladVaidyanathan The direct proof is too hard. For $p>2$ you readily find a nontrivial conjugation $\endgroup$ – Hagen von Eitzen Jul 28 '15 at 10:01

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