# Group of order 1225 is abelian

A problem asked me to show that a group, $G$, of order $1225= 5^{2}\times 7^{2}$ is abelian. I did this by using Sylow's Theorem to show that we have two normal subgroups: one of order $49$ and the other of order $25$. Since the intersection of these subgroups must be $1$, it follows that $G$ is a direct product of these two subgroups. But subgroups of order $p^{2}$ are abelian, where $p$ is a prime. Thus $G$ is abelian.

The question I have upon completing this proof is that it seems to imply that $G$ is isomorphic to $Z_{25} \times Z_{49}$. But this isn't always that case since $Z_{35} \times Z_{35}$ is also a group of order $1225$. Where is my argument failing?

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You haven't given any argument that $G$ is isomorphic to $Z_{25} \times Z_{49}$. If you gave one, maybe we would be able to find the error. – Chris Eagle Aug 19 '12 at 18:30
There are two abelian groups of order $p^2:~\Bbb Z/p\Bbb Z\times\Bbb Z/p\Bbb Z$ and $\Bbb Z/p^2\Bbb Z$. – Olivier Bégassat Aug 19 '12 at 18:31
For completeness, you also have $\mathbb Z_5 \times \mathbb Z_{245}$ and $\mathbb Z_7 \times \mathbb Z_{175}$ – Mark Bennet Aug 19 '12 at 19:31

A group of order $p^2$, where $p$ is a prime, is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_{p} \times \mathbb{Z}_{p}$. You seem to miss the latter case.