Take the 2-minute tour ×

Let $G$ be a finite group. $H \trianglelefteq G$ with $\vert H \vert = p$ the smallest prime dividing $\vert G \vert$. Show $G = HC_G(a)$ with $e \neq a \in H$. $C_G(a)$ is the Centralizer of $a$ in $G$.

To start it off, I know $HC_G(a)\leq G$ by normality of $H$ and subgroup property of $C_G(a)$. So I made the observation that

$\begin{align*} \vert HC_G(a) \vert &= \frac{\vert H \vert \vert C_G(a) \vert}{\vert H \cap C_G(a) \vert}\\&=\frac{\vert H \vert \vert C_G(a) \vert}{\vert C_H(a)\vert}\end{align*}$

But, from here on I never reach the result I'm looking for. Any help would be greatly appreciated!

Note: I posted a similar question earlier, except that one had the index of $H$ being prime, this has the order of $H$ being prime: Link

share|improve this question
You want $a$ to be nonidentity. Also you will in fact just get $G = C_{G}(a)$. Hint for soluton: $a$ has fewer than $p$ conjugates. –  Geoff Robinson Nov 5 '12 at 9:07
Ok, that makes sense, I also got $\vert HC_G(a) \vert = \vert C_G(a) \vert$. And I see what you mean, it follows that $C(a) = \{a\}$ but that one of the things I'm trying to show after proving this property. –  Robert Cardona Nov 5 '12 at 9:11
Well, all conjugates of $a$ are in $H$, so there are less than $p$ of them as $a \neq 1_{G}.$ . On the other hand, their number is $[G:C_{G}(a)]$ which is a divisor of $|G|$. Since $p$ is the smallest prime divisor of $|G|,$ we must have $[G:C_{G}(a)] = 1.$ –  Geoff Robinson Nov 5 '12 at 18:40

1 Answer 1

up vote 1 down vote accepted

Since $N_G(H)/C_G(H)$ injects in $Aut(H)\cong C_{p-1}$ and $p$ is the smallest prime dividing $|G|$, it follows that $N_G(H)=C_G(H)$. But $H$ is normal, so $G=N_G(H)$ and we conclude that $H \subseteq Z(G)$. In particular $G=C_G(a)$ for every $a \in H$.

share|improve this answer

Your Answer


