# Normal subgroup of order $p$ is in the center

Prove that in a group of order $p^2$ ($p$ a prime), a normal subgroup of order $p$ lies in the center. (Since this is Exercise 2.9.6 from Herstein's Topics in Algebra, there are some restrictions on what I can use. This section is on Cayley's theorem. I can show, using the class equation, that $Z(G)>1$ and $G$ is abelian, but I don't want to use the class equation yet.)

If $H$ is a subgroup of order $p$, since $|G|$ does not divide $i(H)!$, $H$ contains a nontrivial proper subgroup $K$ of $G$. Since $p$ is a prime, $K$ must equal $H$. Hence $H$ is normal. Why is it in the center?

• Let $G$ act on $H$ by conjugation; count the fixed points. – user641 Mar 2 '13 at 11:57

This type of exercises are difficult to judge in a forum like this, because we need to know precisely, what has been covered up to that point. Many posters are then inclined to use extra bits they have learned (later in a similar course).

I get the feeling that the following is what might have been expected. This is just fleshing out the hint in Steve D's comment, so I make it a CW.

Let $H$ be a normal subgroup of order $p$. We know (Lagrange's theorem) that $H$ is cyclic. Let $g$ be a generator. Let $x$ be any element of $G$. By normality of $H$ we know that $$xgx^{-1}=g^k,$$ for some integer $k, 0<k<p$. Conjugation by $x$ is an automorphism of $H$, so $xg^{t}x^{-1}=g^{tk}$ for all integers $t$. In particular we get that $$x^2g x^{-2}=x(xgx^{-1})x^{-1}=xg^kx^{-1}=g^{k^2}.$$ An obvious induction then proves that $$x^tgx^{-t}=g^{k^t}$$ for all natural numbers $t$. But, again by Lagrange's theorem $x^{p^2}=1$. Therefore $$g=1g1^{-1}=g^{k^{p^2}}.$$ As $g$ is of order $p$, this means that $1\equiv k^{p^2}\pmod p$. But two applications of Little Fermat tell us that $$k\equiv k^p \equiv k^{p^2}\equiv 1\pmod p.$$ Recalling the constraint $0<k<p$ we can conclude that $k=1$. Therefore $x$ and $g$ commute. Obviously then $x$ commutes with all the powers of $g$. As $x$ was arbitray, we have shown that $H\le Z(G)$.

• I see that your answer is quite delicate and marvellous, but I fail to perceive why this answer flashed out from the comment of Steve D. Per chance some explanations could be provided? Thanks in advance. – awllower Mar 4 '13 at 10:35
• Well, I was just looking at the conjugate action. Something tells me that Steve D's suggestion can lead to a simpler argument, too. Gotta rush now. – Jyrki Lahtonen Mar 4 '13 at 12:59
• This is actually slightly different than what I suggested. Instead, if $G$ acts on $H$ via conjugation, you have a group of order $p^2$ acting on a set of size $p$. So $G$ is being mapped into $S_p$, where each element of the image has order dividing $p^2$. But $G$ fixes one point of $H$ (the identity), and so it must fix at least $p$ points... – user641 Mar 4 '13 at 19:37
• Well, so it seems that the $2$ in the power of $p$ is not of any significance.. So will this hold for any $p-$group? – Qwerty Sep 22 '16 at 21:33
• Well spotted @Qwerty ! It certainly looks like the argument survives. So a normal subgroup of order $p$ of any (finite) $p$-group is in the center. – Jyrki Lahtonen Sep 23 '16 at 4:29

If $H\not\subset Z$ then $H\cap Z=1$. Hence $HZ$ is subgroup of order $\ge p^2$, i.e. $HZ=G$ and $G$ is Abelian.

• Why is $|HZ| \ge p^2$? This follows if $|Z| \ge p$, which hasn't been shown. Why is $G$ abelian? – 64647a Mar 2 '13 at 13:12
• $HZ$ is a $p$-group, so its order is $\ge p^2$. Since $H$ is Abelian and $Z$ is central, so $HZ=G$ is Abelian. – Boris Novikov Mar 2 '13 at 13:16
• $p$-groups aren't defined yet. Can a $p$-group have order $p$? – 64647a Mar 2 '13 at 13:38
• @user64647, but you said you can show $\,|Z(G)|>1\,$ so it must be $\,|Z(G)|=p\,,\,p^2\,$...! – DonAntonio Mar 2 '13 at 13:50
• @DonAntonio: I said I can prove $|Z(G)|>1$ if I used the class equation. Is there another way to prove this? Just asking, because (as I mentioned above) this exercise on $H \le Z$ is asked in a section before the class equation. – 64647a Mar 2 '13 at 13:55

In fact it must be $\,Z(G)=G\Longleftrightarrow G\,$ is abelian, in this case. You only need the easy

Lemma: For any group $\,G\,$ , the quotient $\,G/Z(G)\,$ cannot be cyclic non-trivial.

The above says that $\,G/Z(G)\,\,\,\text{cyclic}\;\Longleftrightarrow \;G\,$ is abelian...