# Fundamental theorem of finitely generated abelian groups.

If $G :=\langle x,y,z \ | \ 2x+3y+5z = 0\rangle$ then find what group $G$ is isomorphic to.

I think I'm supposed to use the fundamental theorem of finitely generated abelian groups, but I don't know if I am using it correctly below.

Since $2x = 0$ we know that $H_x<G$ is a subgroup isomorphic to $\mathbb{Z}_2$ and the same can be said where $H_y\cong \mathbb{Z}_3$ and $H_z\cong\mathbb{Z}_5$. Does this imply that $$G\cong \mathbb{Z}_2\oplus\mathbb{Z}_3\oplus\mathbb{Z}_5$$

Is the logic here correct? Or is there something else I should be doing.

More generally is $\langle x,y,z \ | \ lx+my+nz = 0\rangle \cong \mathbb{Z}_l\oplus\mathbb{Z}_m\oplus\mathbb{Z}_n$ if $l,m,n$ are coprime.

• But $2x \ne 0$. – Derek Holt Nov 5 '14 at 0:01
• $G$ contains the element $(n,n,-n)$ for all $n \in \mathbb{Z}$. So already $G$ contains a subgroup isomorphic to $\mathbb{Z}$. – user452 Nov 5 '14 at 0:12
• $G$ would be that direct sum if $G = \langle x, y, z | 2x = 3y = 5z = 0 \rangle$. As trb456 has pointed out, there's already one free generator; are there two? – Simon S Nov 5 '14 at 0:25
• No that's the only generator. Then to say more about $G$, we need more information about its generators, right? – ItsNotMeItsYou Nov 5 '14 at 0:29

## 2 Answers

Normally you would be taught how to solve problems like this, and shown similar examples.

I can tell you the answer, but I am unsure how much that will help! Since $2$, $3$ and $5$ have no common factor, $2x+3y+5z$ is part of a free generating set of ${\mathbb Z}^3$, so $G \cong {\mathbb Z}^3/{\mathbb Z} \cong {\mathbb Z}^2$.

You could check that $2x+3y+5z$, $x$, $y+2z$ (for example) generates ${\mathbb Z}^3$ (and hence must form a free generating set) by showing that you can express $x$, $y$, and $z$ as linear sums of these three vectors.

This is not a rigorous answer, but it provides some intuition.

Per my comment, we can get $(1,1,-1)$ as a generator of a subgroup isomorphic to $\mathbb{Z}$.

We can also get $(4,-1,-1)$ as another generator of a subgroup isomorphic to $\mathbb{Z}$, and it should be clear that these two generators are independent (i.e. they cannot generate the other).

We can also get $(1,-4,2)$ as another generator of a subgroup isomorphic to $\mathbb{Z}$, except we now have $(4,-1,-1)-3\times(1,1,-1)=(1,-4,2)$. So this third generator can be generated by the prior two.

This is nothing more than an application of the fact that if we have an equation in 3 variables, then if we know two of them we can solve for the third (this is where the 3 coefficients being relatively prime comes into play).

@Derek Holt's answer is telling you that you only have two independent free generators. His answer is the one to select, but perhaps this answer shows what's going on somewhat more concretely.