# 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. Add to solve later

## Hint.

Review the following problems.

## 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 $\langle x, y \rangle \cong \langle x \rangle \times \langle y \rangle$.

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.

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. Add to solve later

### 2 Responses

1. 01/06/2017

[…] order of $P$ is $9$, a square of a prime number, thus $P$ is abelian. (See A group of order the square of a prime is abelian.) Also, the order of the quotient group $G/P$ is $2$, thus $G/P$ is an abelian (cyclic) group. Thus […]

2. 08/12/2017

[…] these Sylow subgroups are of order $11^2$ and $13^2$, respectively, they are abelian. Since the direct product of abelian groups is abelian, the group $G$ is […]

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