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.

Is there any non-abelian group of order $n=49$? I assume there should be at least one but I cannot find an example.

share|improve this question
4  
Actually any group of order $p^2$ for $p$ a prime is abelian. This is a nice exercise using the class equation. –  Qiaochu Yuan Feb 25 '12 at 0:17
1  
$49 = 7^2$. For a given prime $p$ there are only two groups of order $p^2$, and they are both abelian. You should try proving this! –  Dylan Moreland Feb 25 '12 at 0:18
    
thank you for the answers –  the code Feb 25 '12 at 0:26
1  
I took the liberty of removing the boldface text... that seems completely unnecessary. –  Tyler Feb 25 '12 at 2:39

1 Answer 1

up vote 4 down vote accepted

The only groups of order $49$ are $\mathbb Z_{49}$ and $\mathbb Z_7\times \mathbb Z_7$ since $49$ is a square of a prime number.

share|improve this answer

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.