TITLE: Prove that $a^p = e$ in any non-cyclic group of order $p^2$
QUESTION [8 upvotes]: Hey all,
I have an abstract algebra exam coming up and I am unsure of how to do this proof.
Suppose $G$ is a non-cyclic group of order $p^2$, where $p$ is prime.
Prove that $a^p = e$ ($e$ is the identity of the group) for each $a \in G$.
Any help would be appreciated.
Thanks

REPLY [6 votes]: Since you are actually studying for a test, I guess it wouldn't hurt to know this. You can say a little bit more about the group $G$. In general, if $p$ is a prime number and $G$ is a group of order $p^2$, then $$G \cong \mathbb{Z}/p^2 \mathbb{Z} \quad \text{or} \quad G \cong \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}$$
Thus in particular every group of order $p^2$ is abelian and in your case, since you assume that the group is not cyclic, you have $G \cong \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}$ so in this case it is immediate (I hope =P) that every element $a \in G$ satisfies $a^p = e$. You can see a short proof of this here.