# Prove that $a^p = e$ in any non-cyclic group of order $p^2$

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

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Suppose $n$ is the order of $a$. By Lagrange's theorem, $n\mid p^2$. So $n\in\{1,p,p^2\}$. But since $G$ is not cyclic $n\neq p^2$. Hence $n=1$ or $n=p$. In any case we have $a^p=e$.

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Hint: The order of an element must divide the order of the group. (Lagrange's theorem)

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HINT $\rm\ \ a^{n}\: =\ a^{\:p^2}\: =\ 1\ \ \Rightarrow\ \ a^{(n,\ p^2)}\: =\ 1$

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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.

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