[Math] Proving that if $G$ is a group of order $p^2$ where $p$ is prime, then $G$ is either $\mathbb{Z}_p \times \mathbb{Z}_p$ or $\mathbb{Z}_p$

finite-groupsgroup-theory

Prove that if $G$ is a group of order $p^2$ where $p$ is prime, then $G$ is either $\mathbb{Z}_p \times \mathbb{Z}_p$ or $\mathbb{Z}_{p^2}$

I'm aware that this question has an answer here : There exists only two groups of order $p^2$ up to isomorphism. my question however is regarding specific steps used in proofs in the answers of the this question.

So the general idea to prove that $G$ is isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$ or $\mathbb{Z}_{p^2}$ is the following.

  • If $G$ is cyclic, then since $G$ is finite it is isomorphic to $\mathbb{Z}_{p^2}$.

  • If $G$ is not cyclic then by Lagrange's theorem every non-identity element has order $p$ in $G$.

    1. Choose distinct $a\in G, b \in G \setminus \langle a \rangle$ then $\langle a \rangle \cap \langle b \rangle = \{1_G\}$
    2. Both $\langle a \rangle$ and $\langle b \rangle$ are normal in $G$
    3. $\langle a \rangle \langle b \rangle = G$
    4. We can then conclude that $G \cong \langle a \rangle \times \langle b \rangle \cong \mathbb{Z}_p \times \mathbb{Z}_p$

Now I fail to see why both $\langle a \rangle$ and $\langle b \rangle$ are normal in $G$.

The two ways that I could possibly show that $\langle a \rangle$ is normal for example is to show that $g\langle a \rangle g^{-1} \subseteq \langle a \rangle$ for any $g \in G$ or that $|G / \langle a \rangle | = 2$ and in both cases I'm unable to make any progress.

How does one prove that both $\langle a \rangle, \langle b \rangle$ are normal in $G$?

Best Answer

Any group of order $p^2$ (when $p$ is a prime) is abelian. To see this, look at the center of a group $G$ of order $p^2$. It is known that the center of a $p$-group is non trivial, so the center must have order $p$ or $p^2$. If it has order $p^2$ then $G$ is abelian. Suppose it has order $p$. Then we get that $|G/Z(G)|=p$ and hence the quotient group $G/Z(G)$ is cyclic. I'll leave as an easy exercise to prove that $G/Z(G)$ being cyclic implies that $G$ is abelian. But that is a contradiction because if $G$ is abelian then the center must be the whole group and can't have order $p$. Hence $|Z(G)|=p^2$ and $G$ is abelian. And of course any subgroup of an abelian group is normal in it.

Related Question