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I need to show if $|G|=p^2$, then either $G\cong \mathbb{Z}_{p^2}$ or $G\cong \mathbb{Z}_{p} \times \mathbb{Z}_{p}$.

Let $x$ be any element in $G$. Then $o(x)$ divides $p^2$. There are three possible cases: either $o(x)=1$, or $o(x)=p$ or $o(x)=p^2$. If $o(x)=1$, then $x=e$. Now, if $o(x)=p^2$, then $G=\langle x \rangle$, thus $G \cong \mathbb{Z}_{p^2}$.

And now this is part where I get stuck. If $o(x)=p$ then every element in the group has an order $p$. Let $a,b \in G$ with $a,b \ne e$ and $b$ is not a power of $a$. Now I need to show $\langle a \rangle \cap \langle b \rangle = \{ e \}$. Suppose $x\in \langle a \rangle \cap \langle b \rangle $. Then $x=a^m$ and $x=b^n$ for some integers $m,n$. Thus, $a^m=b^n$. I'm stuck here. How do I show that the previous condition forces $x$ to be the identity? I can prove the rest if this holds true.

Also, I know that if $|G|=p^2$ then $G$ is abelian.

ash
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    https://math.stackexchange.com/questions/317356/there-exists-only-two-groups-of-order-p2-up-to-isomorphism?noredirect=1&lq=1 – Lucas Mar 01 '18 at 17:37

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The key point is the $\langle a \rangle \cap \langle b \rangle$ is a subgroup of both $\langle a \rangle$ and $\langle b \rangle$, and these have order $p$, and so have no proper subgroup.

lhf
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