I just proved that any finite group of order $p^2$ for $p$ a prime is abelian. The author now asks to show that there are only two such groups up to isomorphism. The first group I can think of is $G=\Bbb Z/p\Bbb Z\oplus \Bbb Z/p\Bbb Z$. This is abelian and has order $p^2$. I think the other is $\Bbb Z/p^2 \Bbb Z$.
Now, it should follow from the fact that there is only one cyclic group of order $n$ up to isomorphism that these two are unique up to isomorphism. All I need to show is these two are in fact not isomorphic. It suffices to show that $G$ as before is not cyclic. But this is easy to see, since we cannot generate any $(x,y)$ with $x\neq y$ by repeated addition of some $(z,z)$.
Now, it suffices to show that any other group of order $p^2$ is isomorphic to either one of these two groups. If the group is cyclic, we're done, so assume it is not cyclic. One can see that $G=\langle (1,0) ,(0,1)\rangle$. How can I move on?
Best Answer
Let $G$ be a $p^2$ group. As you said, it is Abelian.
Note that the order of every element divides $p^2$, so is equal to $1$ (for the identity $e$ only), $p$, or $p^2$.
If there is an element $x$ of order $p^2$, then $G=\langle x\rangle$ by cardinality. So $G$ is cyclic, and as you pointed out $$ G\simeq \mathbb{Z}/p^2\mathbb{Z}. $$
Now assume that there is no element of order $p^2$. This means that every element which is not the identity has order $p$. Pick $x$ order $p$. Since $\langle x \rangle\subsetneq G$, you can take another order $p$ element $y$ in the complement of $\langle x \rangle$.
Now $$ \theta:(u,v)\longmapsto uv $$ yields a homomorphism from $\langle x \rangle\times\langle y \rangle$ to $G$. Note that $\langle x\rangle\cap\langle y\rangle=\{e\}$, so the latter is injective. Since both groups have the same cardinality $p^2$, it follows that $\theta$ is an isomorphism.
Finally, since $\langle x\rangle \simeq\langle y\rangle \simeq \mathbb{Z}/p\mathbb{Z}$, we have $$ G\simeq \langle x \rangle\times \langle y \rangle\simeq \mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}. $$
So $G$ is either isomorphic to $ \mathbb{Z}/p^2\mathbb{Z}$ or to $\mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$.