Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
6  
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
add comment

1 Answer

up vote 10 down vote accepted

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.

share|improve this answer
    
Thanks! I realize my mistake. –  neelp Aug 19 '12 at 19:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.