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Let $G$ be a finite group. Let $H \trianglelefteq G$, with $\vert H \vert = p$, a prime, where $p$ is the smallest prime dividing $\vert G \vert$. Prove that $H \leq Z(G)$. (Hint: If $a \in H$, by normality, its conjugacy class lies inside $H$.)

My approach so far:

Let $a \in H$. Then $C(a) = \{gag^{-1} : g \in G\} \subseteq H$. Now, since $\vert H \vert$ is prime, it follows that $H$ is cyclic and moreover $H$ is abelian, hence $C_H(a) = H$ for all $a \in H$ and equivalently $[H: C_H(a)] = 1$ for all $a \in H$.

Try to show $\vert C(a) \vert = [G:C_G(a)] = [H:C_H(a)] = 1$.

So, since $\vert C(a) \vert = 1$ for all $a \in H$, we have $C(a) = \{a\}$ for all $a \in H$ and equivalently $a \in Z(G)$ for all $a \in H$, so we can say $H \subseteq Z(G)$ and moreover $H \leq Z(G)$.

My main question is, how to show $[G:C_G(a)] = [H:C_H(a)]$. But also, I haven't used the fact that $p$ is the smallest prime dividing $\vert G \vert$, so is my reasoning wrong anywhere?

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3 Answers 3

up vote 3 down vote accepted

We already know $e \in H$ is also in the center of the group.

Now pick $a \in H \backslash \{e\}$.

You know $C(a)$ is a subset of $H$, so that $|C(a)| \leq |H| = p$.

We also know $|C(a)|$ divides $|G|$.

But $p$ is the smallest prime dividing $|G|$.

Thus, either $|C(a)|$ is $p$ or $1$.

If $|C(a)| = p$, then $C(a) = H$. In particular, we have $e \in C(a)$, hence $gag^{-1} = e$ for some $g \in G$.

But then $a = e$; contradiction.

Thus, $|C(a)| = 1$, as you wanted.

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Because the centralizer of an element of a group is a subgroup of the group (proof?) –  Chasky Nov 3 '12 at 5:40
    
$C(a) = \{gag^{-1} : g \in G\}$. In this case, for all $e \neq a \in H$, we will have $C(a) = \{a\}$ as a result of this statement I'm trying to prove. Correct? So its NOT a subgroup most of the time? –  Robert Nov 3 '12 at 5:44
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@Rankeya because $\vert C(a) \vert = [G: C_G(a)]$, By Orbit-Stabilizer Theorem –  Robert Nov 3 '12 at 5:47
    
Okay. I am deleting my earlier comments, because there have been edits to the question. –  Rankeya Nov 3 '12 at 5:48
    
Of course Robert. Sorry, I was still thinking about $C(a)$ being a subgroup thing. Confused me. Answer looks good now. –  Rankeya Nov 3 '12 at 5:51
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We use the following fact.

If $G$ is group and $H \le G$ , $N_G(H)/ C_G(H) $ is isomorphic to a subgroup of $\text{Aut}(H)$. ($N_G(H)$ is the normalizer of $H$ and $C_G(H)$ is the centralizer of $H$ in $G$. )

Since $p$ is a prime, $H$ is cyclic. So $|\text{Aut}(H)| = \varphi(p) = p-1 $. So by Lagrange's theorem $|N_G(H)|/ |C_G(H)|$ divides $p-1 $. Since $N_G(H) = G$, $\frac{|G|}{C_G(H)}$ divides $p-1$. Since $p$ is the smallest prime dividing $G$, we must have $|C_G(H)| = |G|$.So $G = C_G(H)$. This is same as saying $H \le Z(G)$.

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Since $H\trianglelefteq G$ then $H=\bigcup_{x\in H}C(x)$ so for all $x,y\in H$ $$C(x)\cap C(y)=\emptyset \; \; \; \text{or} \; \; \;C(x)=C(y)$$ Moreover $|H|=\sum_{x\in H}|C(x)|$ such that $[G:C_G(H)]=|C(x)|\bigg||G|$ as you noted. Since $p$ is a smallest prime dividing the order of the group then we have $$|C(x)|=1, \; \; \; \forall x\in H$$ as desired. (-;

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+1! ;^) dear friend! –  amWhy Apr 3 '13 at 0:58
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