# $P\in \operatorname{Syl}_p(S_n)$ implies that $P\in \operatorname{Syl}_p(A_n)$ and $|N_{A_n}(P)|=\frac{1}{2}|N_{S_n}(P)|$

$\newcommand{\Syl}{\operatorname{Syl}}$

This is an exercise (with hint about the second part) in my own language book in Group theory, however, maybe it is a lemma or theorem in an standard book which I don't have. It is:

Let $p$ is an odd prime number and $n$ is a natural number such that $p\leq n$. Therefore if $P\in \Syl_p(S_n)$ then $P\in \Syl_p(A_n)$ and $|N_{A_n}(P)|=\frac{1}{2}|N_{S_n}(P)|$.

What I have done: Suppose that $P\in \Syl_p(S_n)$ where in $S_n$ is the symmetric group on $n$ letters. For having $P\in \Syl_p(A_n)$, I clearly should prove $P\leq A_n$ first. I can see no clues here to prove that unless I assume $P\nleq A_n$. With this assumption, $P$ has an odd permutation. The rest is unclear to me. :( Maybe, I am losing some obvious facts for proving the first part?

For the second part that is $|N_{A_n}(P)|=\frac{1}{2}|N_{S_n}(P)|$; there is a hint noting use the Frattini Argument. I can handle this part. :) Indeed, I see $S_n=N_{S_n}(P)A_n$ and the rest is routine. Thanks for helping me about the first part.

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If $|P|=p^k$, check whether $p^k$ divides $A_n$ as well. $A_n$ being normal means if it contains one Sylow subgroup, then it contains all. –  user641 Aug 20 '12 at 16:48

## 2 Answers

Suppose $|P| = p^k$ and $n! = p^k m$. Since $n!$ is divisible by 2 and $p$ is odd, $m$ is divisible by 2. Let $m' = m/2$. Then $n!/2 = p^k m'$. Since $m'$ is not divisible by $p$, there exists $Q \in Syl_p(A_n)$ such that $|Q| = p^k$. Since $Q \in Syl_p(S_n)$, $P$ and $Q$ are conjugate. Since $A_n$ is a normal subgroup of $S_n$, $P \in Syl_p(A_n)$.

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Great! Wonderful! –  Babak S. Aug 21 '12 at 10:03

Makoto Kato (and Steve D in a comment) gave an argument based on calculating the orders of the respective Sylow $p$-subgroups (+1). An alternative approach is to observe that any element $x\in P$ has an order that is a power of $p$. Therefore the same holds for all its cycles, so they all have an odd length. Hence $x\in A_n$, and $P\subseteq A_n$.

As you observed a Frattini argument (or a counting argument) shows that the normalizer $N_{S_n}(P)$ must contain an odd permutation, and the claim about the orders of the normalizers follows from this.

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Thanks for your another approach Jyrki. I hope to think like you in this approach some day.Still learning great facts in Group Theory. Thanks :) –  Babak S. Aug 21 '12 at 10:06