A Group of Order the Square of a Prime is Abelian

Problem 20

Suppose the order of a group $G$ is $p^2$, where $p$ is a prime number.
Show that

(a) the group $G$ is an abelian group, and

(b) the group $G$ is isomorphic to either $\Zmod{p^2}$ or $\Zmod{p} \times \Zmod{p}$ without using the fundamental theorem of abelian groups.

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Hint.

Review the following problems.

1. The center of a p-group is not trivial (post 1)
2. If the quotient by the center is cyclic, then the group is abelian (post 2)

Proof.

(a) A group of order $p^2$ is abelian.

Since $G$ is a $p$-group, its center is not trivial (see post 1   for a proof.)

If the center $Z(G)=G$, then $G$ is abelian so assume that $Z(G)$ is a proper nontrivial subgroup. Then the center must have order $p$ and it follows that the order of the quotient $G/Z(G)$ is $p$, hence $G/Z(G)$ is a cyclic group.

Thus $G$ is abelian by the fact proved in post 2.

(b) The group $G$ is isomorphic to either $\Zmod{p^2}$ or $\Zmod{p} \times \Zmod{p}$

Let $x \in G$ be any nontrivial element of $G$. If $x \in G$ has order $p^2$, then $G=\langle x \rangle \cong \Zmod{p^2}$. If the order of $x$ is $p$, then take $y \in G \setminus \langle x \rangle$. Then the order of $y$ is also $p$.

Since the subgroup $\langle x, y \rangle$ generated by $x$ and $y$ is properly bigger than the subgroup $\langle x \rangle$, we must have $G=\langle x, y \rangle$.
We claim that $\langle x, y \rangle \cong \langle x \rangle \times \langle y \rangle$.

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Define a map $f:\langle x \rangle \times \langle y \rangle \to \langle x, y \rangle$ by sending $(x^a, y^b)$ to $x^ay^b$. This is a group homomorphism because for any elements $(x^{a_1}, y^{b_1})$ and $(x^{a_2}, y^{b_2})$, we have
\begin{align*}
f\left( (x^{a_1}, y^{b_1})(x^{a_2}, y^{b_2}) \right) &=f\left (x^{a_1+a_2}, y^{b_1+b_2})\right) =x^{a_1+a_2} y^{b_1+b_2}\\
& =x^{a_1}y^{b_1}x^{a_2}y^{b_2} =f\left( (x^{a_1}, y^{b_1}) \right) f\left( (x^{a_2}, y^{b_2}) \right).
\end{align*}

Here we used the result of part (a) that $G$ is abelian in the third equality.

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We claim that the homomorphism $f$ is injective.
If $f\left (x^a, y^b) \right)=1$, we have $x^a=y^b$ but since $y \not \in \langle x \rangle$ we must have $a=b=0$. Thus the kernel is trivial, hence $f$ is injective.

Since $\langle x \rangle \times \langle y \rangle \cong \Zmod{p} \times \Zmod{p}$ has order $p^2$ and $f$ is injective, the homomorphism must be surjective as well, hence it is an isomorphism.

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