Number of non-isomorphic groups of order $p^2$

Question

The number of non-isomorphic groups of order $p^2$, where $p$ is a prime number is:

1. 1

2. $p$

3. 2

4. $p^2$

What is simplest method to find number of non-isomorphic group? I read from various books but I don't understand the method.

I know that a finitely generated abelian group is isomorphic to direct product of cyclic group.

Answer 1

First, note any group of order $p^2$ is abelian. If $|Z(G)|=p^2$, this is automatic. If $|Z(G)|=p$, then $G/Z(G)$ has order $p$, hence is cyclic, so $G$ is abelian (a well known fact), hence actually $|Z(G)|=p^2$. So necessarily we must be in the first case. Analyzing these two cases suffices since $p$-groups have nontrivial center.

So any group of order $p^2$ is a finitely generated abelian group, hence is a direct product of cyclic subgroups by the structure theorem. The only possibilities are $\mathbb{Z}/(p^2)$ or $\mathbb{Z}/(p)\times\mathbb{Z}/(p)$, so there are only two groups of order $p^2$, up to isomorphism.

Answer 2

Hint: Suppose $G$ is an Abelian group with $|G|=p^2$(which is always the case, as Ben has shown above). There can be three possibilities of order of $a\in G$

$|a|=1,p \text{ or }p^2$

Case I: There is an element or order $p^2$. In this case $G$ is cyclic and $G\cong \mathbb{Z}_{p^2}$

Case II: All non-identity elements are of order $p$

In this case $G≃\mathbb{Z}_p\times\mathbb{Z}_p$

To show this, pick up an element $x$ of order $p$ and $y∈G−⟨x⟩$ and then prove $G≃G/⟨x⟩×G/⟨y⟩≃ \mathbb{Z}_p\times\mathbb{Z}_p$

Thus, there are two non-isomophic groups of order $p^2$.