TITLE: Groups of order $25$
QUESTION [6 upvotes]: Please verify my solution that there are only two groups of order $25$ up to isomorphism.
As $|G|$ is a prime squared, then $G$ is abelian.
Since the Theorem of Finite Abelian Groups, $G$ is a direct product of cyclic groups. The only possibilities here, since $25=5.5$, are $G = \mathbb{Z_{25}}$ or $G = \mathbb{Z_5} \times \mathbb{Z_5}$. Note that there is no element or order $25$ in the latter, so they're not isomorphic.

REPLY [4 votes]: Your solution is correct,here is a solution without using the Structure theorem.
Proof: Suppose $G$ is cyclic then $G \cong \frac {\mathbb Z}{25 \mathbb Z}$. So assume that $G$ is not cyclic and $G$ is abelian (as you mentioned in your solution).Now note that we have a group action $\frac {\mathbb Z}{5 \mathbb Z} \times G \to G$ defined as $(a,g) \to g^a$ (prove yourself that its actually a group action!).Hence $G$ is a vector space of dimension $2$ over $\frac {\mathbb Z}{5 \mathbb Z}$,therefore is isomorphic to $\frac {\mathbb Z}{5 \mathbb Z} \times \frac {\mathbb Z}{5 \mathbb Z}$. QED