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.



the number of Sylow subgroups
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.


