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?

Solution 1:

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 $H\le Z(G)$.

Solution 2:

If $G$ were cyclic it would be abelian, and we’d be done. Suppose it’s not. Then every element is of order $p$ by Lagrange’s theorem. And every subgroup of size $p$ is normal by Herstein’s lemma 2.9.1.

Subgroup $H$ must be cyclic having elements of order $p$. Then any two elements of $H$ commute.

Suppose $k\not\in H$. Then $k$ generates a group $K$ of size $p$, and this group is normal (using the first paragraph).

Now $H\cap K=\{e\}$ since it’s a subgroup of size $<p$. Then, for any $h\in H,$ $hkh^{-1}k^{-1} \in H\cap K$ using the normalcy of both subgroups (eg $hkh^{-1}\in K$ since $K$ is normal, and so on). But then $$hkh^{-1}k^{-1} = e \Rightarrow hk = kh.$$

Solution 3:

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.