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.


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 $N_G(H) = C_G(H)$.

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

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