Let $G$ be a group of order $p(p^2+1)$, where $p$ is an odd prime number and $p>3$. Easily we can see that $G$ is solvable and so $G$ has a Hall subgroup $L$ of order $p^2+1$. Also we know that $P$, the Sylow subgroup of $G$ is a normal subgroup of $G$. Is it true in general that $L$ is a normal subgroup of $G$?
As Jeremy Rickard points out, the answer is no in general. However, $G$ has order divisible by $2,$ but not by $4.$ It follows that $G$ has a normal $2$complement $K$ of order $p \left( \frac{p^{2}+1}{2} \right).$ Now $K$ has a normal $p$complement by Burnside's transfer theorem, since ${\rm gcd}(p1,\frac{p^{2}+1}{2}) = 1.$ This subgroup of $K$ has order $\frac{p^{2}+1}{2}$ and is characteristic in $K,$ hence normal in $G.$ Thus there is a subgroup of index $2$ in $L$ which is normal in $G.$

1$\begingroup$ I guess a slightly easier argument, using the normality of $P$, is: $L/\text{C}_L(P)$ is a subgroup of $\text{Aut}(P)=C_{p1}$. But $\text{gcd}(p1,\lvert L\rvert)=2$, so $[L:\text{C}_L(P)]\le2$. $\endgroup$ – Peter Mueller May 8 '14 at 13:56

$\begingroup$ Yes, marginally easier. I had forgotten that $P$ was normal, so Burnside's transfer theorem certainly wasn't necessary. $\endgroup$ – Geoff Robinson May 8 '14 at 13:57
No. For example, take the direct product of a dihedral group of order $2p$ and a cyclic group of order $(p^2+1)/2$.