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.


2 Answers 2


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.

  • $\begingroup$ Many thanks for the very excellent answer. $\endgroup$
    – BHZ
    Aug 5, 2013 at 20:16

Let $k$ be the number of $p$-Sylow subgroups. Then $k$ divides $2(p^2+1)$. Let $l$ be the $2(p^2+1)/k$. $k$ is congruent to $1$ mod $p$, so $l$ is congruent to $2$ mod $p$. So $k=ap+1$ and $l=bp+2$. But $ab\leq 2$ clearly so we can divide into three cases:

$a=1,b=1$: $(p+1)(p+2) \neq 2p^2+2$ unless $p=3$.

$a=2, b=1$ $(2p+1)(p+2) \neq 2p^2+2$

$a=1,b=2$: $(p+1) (2p+2) \neq 2p^2+2$

so each case is impossible, except $a=0$ and $b=0$. If $a=0$ we are done, so we consider the case $b=0$.

The normalizer of a Sylow subgroup is a group of order $2p$, hence it is either $\mathbb Z/{2p}$ or $D_p$. This gives a division into two cases.

For another division into two cases, either this subgroup is a Frobenius complement, or the normalizers of different Sylow subgroups intersect.

This gives $2\times 2=4$ cases. However, the subgroup cannot be a Frobenius complement, because $\mathbb Z/p$ cannot act nontrivially on a group of order $p^2+1$. Since $p^2+1=2$ mod $4$, the Sylow $2$-subgroups are all $\mathbb Z/2$, so the number of elements of order $2$ divides $p^2+1$, so it is not a multiple of $p$, so there is a $p$-invariant element. But the group of $p$-invariant elements also has order dividing $p^2+1$, and is congruent to $1$ mod $p$ (because the order of its complement is a multiple of $p$), so its order is $1$ or $p^2+1$, so the action is trivial.

We can further use this fact to eliminate the case where the group is $\mathbb Z/2p$. Let $e$ be an element shared by the normalizers of two different Sylow $p$-subgroups, then it must have order $2$, and it commutes with those elements, so $Z(e)$ has at least two different Sylow $p$-subgroups. The order of $Z(e)$ divides $2p(p^2+1)$, so by the same argument as the beginning, the number of Sylow subgroups in $Z(e)$ is $p^2+1$, so $Z(e)/(e)$ has order $p(p^2+1)$, so the Sylow $p$-subgroup in that is a Frobenius complement. But this is a contradiction because $\mathbb Z/p$ cannot act nontrivially on a group of order $p^2+1$.

This leaves just one case, when the normalizer of a Sylow $p$-subgroup is $D_p$, and different normalizers do intersect. I do not know how to eliminate this case.

  • 2
    $\begingroup$ The $D_p$ case is not settled yet? $\endgroup$
    – user6976
    Aug 5, 2013 at 6:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.