Q:

Property of Sylow $2$-subgroups of $S_n$ and $S_{n+1}$

The question

Let $P$ be a Sylow $2$-subgroup of $S_n$ and let $Q$ be a Sylow $2$-subgroup of $S_{n+1}$. Show $Q\cong P\times C_2$ iff $n\equiv 1\pmod{4}$.

My attempt
The first direction is obvious enough: if $Q\cong P\times C_2$, $o(Q)=2o(P)$ meaning $n$ must satisfy $n\equiv 1\pmod{4}$, so that $n+1$ contributes exactly one more $2$ to the prime decomposition of $o(S_{n+1})=(n+1)!$. The second direction, however, seems to be much more difficult. Assuming $n\equiv 1\pmod{4}$, and noting that $S_n$ is (clearly) isomorphic to a subgroup of $S_{n+1}$, we can conclude $P$ is isomoprhic to a subgroup of $Q$, by Sylow's second theorem. Now, since $n\equiv 1\pmod{4}$, it is clear $P$ has index $2$ in $Q$ (well, the map of $P$ under the embedding of $S_n$ in $S_{n+1}$) and it is hence normal in $Q$. Next, we also have $o(Q)=o(P)\cdot o(C_2)$ and if we can show $C_2\lhd Q$ and $C_2\cap P=\{1\}$ we'll be done. To show that $C_2\cap P=\{1\}$, I note $C_2\cap P$ is a subgroup in $C_2$ and hence is either $\{1\}$ or $C_2$ itself. In the latter case, $C_2\subseteq P$, which I can't seem to contradict as well.
Questions

How do I show that $C_2\cap P=\{1\}$?
How do I show that $C_2$ is (isomorphic to) a normal subgroup of $Q$?

A:

We have $S_{n-1}\le S_n\le S_{n+1}$ via the obvious embeddings.
Let $R$ be a 2-Sylow of $S_{n-1}$. As $n$ is odd, we see from its order that $R$ is also a 2-Sylow of $S_n$. So may may as well take $P=R$. Note that this $P$ leaves $n$ and $n+1$ fixed. Hence the group generated by $P$ and $(n\,n+1)$ is a subgroup of $S_{n+1}$ isomorphic to $P\times C_2$, and has exactly the order needed for a 2-Sylow of $S_{n+1}$.

