0
$\begingroup$

I am working on the following question: Suppose $G$ is a finite group that has a cyclic 2-Sylow subgroup $H$. I want to show that the centralizer, $C_G(H)$, and $\text{normalizer,} \ N_G(H)$ coincide.

My attempt: Since $H$ is a cyclic 2-Sylow subgroup, $H$ has order $2^k$ for some $k \geq 1$. Also, $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $Aut(H)$. $Aut(H)$ has order $2^{k-1}(2-1)=2^{k-1}$. So, the order of $N_G(H)/C_G(H)$ must be even and of the form $2^j$, where $j\leq k-1$. I need to show that $j$ is $0$ then so that the order is $1$, but I think I am missing something simple to complete this proof.

$\endgroup$
2
$\begingroup$

As $H$ is abelian, $C_G(H)$ contains $H$, that is, a $2$-Sylow subgroup of $N_G(H)$. Hence $N_G(H)/C_G(H)$ has order coprime to $2$, but as you already realized, $N_G(H)/C_G(H)$ is isomorphic to a subgroup of the $2$-group $\text{Aut}(H)$ and so we get $N_G(H) = C_G(H)$.

$\endgroup$
  • $\begingroup$ Oh, that makes sense, thanks! $\endgroup$ – user196559 Dec 16 '14 at 6:26

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.