If $G$ is a group of order $p^2$ with $p$ a prime number then $G \simeq C_{p^2}$ or $G \simeq C_p \times C_p$

Asked Jun 16 '17 at 13:43

Active Jun 16 '17 at 13:43

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Here's my attempt

We have $|G| = p^2$, Therefore, $o(g)= 1$ or $p$ or $p^2 \forall g$ where $o(g)$ denotes the order of the element of $g \in G$

Let's differentiate the $3$ cases.

If $o(g)= 1$ then $g=e$ by unicity of the identity of G
If $o(g) = p^2$ then $\langle g\rangle = G \implies G \simeq C_p^2$
Now it's obvious that I have to show that if $o(g)= p$ then $G \simeq C_p \times C_p$ But I'm not sure how to do that.

asked Jun 16 '17 at 13:43

Are you already familiar with the lemma: "The center of a $p$-group is never trivial"? – Jun 16 '17 at 13:46
(by the way, in the second $\bullet$ it should have been $C_{p^2}$, not $C_p^2$) – Jun 16 '17 at 13:47
See also https://math.stackexchange.com/questions/415825/group-g-of-order-p2-g-cong-mathbb-z-p2-or-g-cong-mathbb-z-p-tim?noredirect=1&lq=1 – Arnaud D. Jun 16 '17 at 15:02

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