Why are Sylow 2-subgroups of $S_n$ self-normalizing?

Why are Sylow 2-subgroups of $S_n$ self-normalizing?

I've read before that this is the case but haven't seen a proof.

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Let $M$ be the subgroup generated by $(12)$, $(34)$, $(56)$, etc., and let $P$ be the Sylow 2-group which contains $M$. $M$ is normal in $P$, and if $g\in S_n$ normalizes $P$, then it normalizes $M$; in other words, $N_G(P)\subseteq N_G(M)$.
Now by counting conjugates of $M$, we see that $N_G(M)$ has order $\lfloor \frac{n}{2}\rfloor!\cdot|M|$. Note that of course $M$ is in $N_G(M)$; also, any permutation on the "odd" points ($1,3,5,\ldots$) can be made into an element normalizing $M$, by simply doing the same thing to the "even" points. This creates a copy of $S_{\lfloor \frac{n}{2}\rfloor}$ (call it $K$) inside $N_G(M)$. Taking orders, we see that $N_G(M)$ is just $M\rtimes K$, and that $P$ is just $MQ$, where $Q$ is a Sylow 2-subgroup of $K\cong S_{\lfloor \frac{n}{2}\rfloor}$.
This means to find the normalizer of $P$ (which remember lives in $N_G(M)$), it suffices to find the normalizer of $Q$ in $S_{\lfloor \frac{n}{2}\rfloor}$. By induction, we're done.
Why is $M$ normal in $P$? And is it actually characteristic? – Cihan Apr 13 '12 at 15:26
Not characteristic. But the generators of $M$ are the only transpositions in $P$. So if $g$ normalizes $P$, it has to send these transpositions to some rearrangement of themselves. – user641 Apr 13 '12 at 16:15