# Group relations and generators

I have to find out what is this abelian group (in the form $\mathbb{Z}/m_1\mathbb{Z} \times ...$). Its relations are: $$6a+9b+6c=0$$ $$8a+12b+4c=0$$ with generator $a,b,c$. My solution is: $$\mathbb{Z}/12\mathbb{Z}\times \mathbb{Z}$$ But the first term should be a $14$ instead according to the book, Can someone help me and explain how to get there? I don't understand it.

My work: I transform the system up there into: $$6t+3u=0$$ $$8t+4u-4v=0$$ with $t=a+b+c$, $u=b$, $v=c$ Then, I transform it again in: $$3x=0$$ $$4x-4\mathbb{Z}=0\longrightarrow 4(x-\mathbb{Z})=0$$ with $x=u+2t$, $v=\mathbb{Z}$ So the group is $$\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}$$ And $\mathbb{Z}_3\times \mathbb{Z}_4$ is isomorphic to $\mathbb{Z}/12\mathbb{Z}$, so the group is: $$\mathbb{Z}/12\mathbb{Z}\times \mathbb{Z}$$

Can someone check that's right?

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Presumably, you mean abelian group – Thomas Andrews Jan 26 '13 at 19:36
Ho did you get $12$ – Thomas Andrews Jan 26 '13 at 19:37
12 is correct.. – user58512 Jan 26 '13 at 19:40
Yes it's abelian. I got the twelve just by transforming the relations and getting new generators, until I had something of the form $zu_1=0$ So I know that corresponds to $Z/xZ$ – MyUserIsThis Jan 26 '13 at 19:44
@anon yes, that was a typo. I wrote what I've done. – MyUserIsThis Jan 26 '13 at 19:59