Sylow subgroups and isomorphism 
Let $P$ a Sylow-2 subgroup of $S_n$. How can we prove that all Sylow-2 subgroups of $S_{n+2}$ isomorphic to $P\times C_2 $ if and only if $n\equiv 0 (\mod 4)$ or $n\equiv 1 (\mod 4)$. 

Could you please help?
 A: Hint - what can you say about the order of the Sylow-$2$ subgroups of $S_{n+2}$ in the various cases of $n$ modulo $4$?
Hidden text below gives some more details of a way through. I haven't done all the work - there are some details to add. If you think about it you should see how to make some tweaks and improvements.

If $n=4m$ then $n+1$ is odd and $n+2=4m+2=2(2m+1)$ is twice an odd number. Hence the order $(n+2)!$ of $S_{n+2}$ has precisely one more factor of $2$ than the order of $S_n$. In the case of $n=4m+1$ the only extra factor of $2$ also comes from $4m+2$.

Hence the Sylow $2-$subgroup of $S_{n+2}$ has order $2|P|$ to reflect the extra factor of $2$. $P\times C_2$ has the right order. Taking $S_m$ as the group of permutations on the first $m$ positive integers, we can use $(n+1\text{ } n+2)$ as a transposition which commutes with everything in $S_n$ to obtain a group of precisely this form. Since all Sylow subgroups for a given prime are conjugate, they all have this order and form. That does the if part.

For "only if" you need to think what happens for $n=4m+2$ and $4m+3$ and the extra factors of $2$ you pick up. You should be able to conclude that although $P\times C_2$ is a $2-$group it is too small to be a Sylow $2-$subgroup.

