Exercise 4.2.9 (Every group of order $p^2$ has a normal subgroup of order $p$)

Proof. If $|G|=p^{\alpha }(\alpha >0)$, then $p$ is the smallest prime dividing $|G|$. By Corollary 5, every subgroup of index $p$ is normal in $G$.

We repeat the proof of Corollary 5 in the particular case $|G|=p^{\alpha }$:

Suppose $H\le G$ and $|G:H|=p$. Let ${\pi }_H$ be the representation afforded by multiplication on the set $A$ of left cosets of $H$ in $G$, let $K=\ker <!--\; FUNCTION\; APPLICATION\; -->({\pi }_H)$ and let $|H:K|=k$. Since $k$ divides $|H|$, and $|H|$ divides $p^{\alpha }$, $k=p^{\beta }$ for some integer $\beta$ with $0\le \beta \le \alpha$, and

Since $H$ has $p$ left cosets, $G∕K\simeq {\pi }_H(G)\le S_A$ by the First Isomorphism Theorem, so $G∕K$ is isomorphic to a subgroup of $S_A\simeq S_p$. By Lagrange’s Theorem, $p\cdot p^{\beta }$ divides $p!$, thus

If $\beta >0$, then $p\mid (p-1)!$. This is impossible, since $(p-1)!\equiv -1(\mathit{𝑚𝑜𝑑}p)$ by Wilson’s Theorem, so $\beta =0$. Hence $|H:K|=1$, so $H=K$ is normal in $G$.

Suppose now that $|G|=p^2$.

• If there is some element $x\in G$ of order $p^2$, then $⟨x⟩\subseteq G$ and $|⟨x⟩|=p^2=|G|$, therefore

  $$G=⟨x⟩.$$

  Then $|x^p|=p$, so $H=⟨x^p⟩$ is a subgroup of $H$ of order $p$, and of index $p$ in $G$.

• Otherwise, since all elements have order $1,p,p^2$, all elements $g\ne 1$ are of order $p$, so there is an element $x$ of order $p$. Then the index of $H=⟨x⟩$ in $G$ is $p$.

So every group of order $p^2$ has a subgroup of index $p$ in $G$.

By the first part, such a subgroup is normal.

In conclusion, every group of order $p^2$ has a normal subgroup of order $p$. □