TITLE: the number of Sylow subgroups
QUESTION [5 upvotes]: Let $p>3$ be a prime number and $G$ be a finite group of order $2p(p^2+1)$. 
Is it true that always the Sylow $p$- subgroup of $G$ is a normal subgroup of $G$?
As I checked by Gap it seems true.
Thanks for your answer.

REPLY [14 votes]: As $p^2+1 \equiv 2 \bmod 4$, a Sylow $2$-subgroup of $G$ has order four.  As $p^2+1 \equiv 2 \bmod 3$ and $p>3$, $G$ has no element of order three.  It follows now that a Sylow $2$-subgroup $X$ of $G$ is central in its normalizer in $G$.  By Burnside's normal $p$-complement theorem, $G$ contains a normal complement $N$ to $X$.  It suffices to show that $N$ has a normal Sylow $p$-subgroup, as such a subgroup is characteristic in $N$ and thus normal in $G$.
Let $t$ be the number of Sylow $p$-subgroups of $N$.  Then $t \equiv 1 \bmod p$ and $t$ divides $\frac{p^2+1}{2}$.  Assume for contradiction that $t>1$, write $t=ap+1$ with $a>0$.  Then $p^2+1=2b(ap+1)$ for some integer $b$.  Now $2b \equiv 1 \bmod p$.  So, $2b>p$ and $2b(ap+1)>p^2+p$, giving the desired contradiction.