Re: generators and relations

I think I've made some progress on this problem.

$\displaystyle 4a+3b+3c=0$

$\displaystyle 2a-b+3c=0$

Subtracting twice the second from the first gives:

$\displaystyle 2a-b+3c=0$

$\displaystyle 5b-3c=0$

And now subtracting the second from the first gives:

$\displaystyle 2a-6b+6c=0$

$\displaystyle 5b-3c=0$

Let $\displaystyle e=a-3b+3c$, so $\displaystyle a=e+3b-3c$ gives:

$\displaystyle 2e=0$

$\displaystyle 5b-3c=0$

The first equation gives $\displaystyle \mathbb{Z}/2\mathbb{Z}$, and for the second, my reasoning is that $\displaystyle \mathbb{Z}\times\mathbb{Z}$ gets divided into cosets of $\displaystyle \{(3x,5x):x\in\mathbb{Z}\}$, and so that gives a factor of $\displaystyle \mathbb{Z}$. So the answer is $\displaystyle M\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}$.

Is that correct?

- Hollywood