How to describe the quotient group Z x Z / < (4, -6)>

While solving a problem on group theory, I encountered the quotient group Z x Z / < (4, -6)>. Here Z is the integer.

At first I thought it is just Z/(4Z) x Z/(6Z). But I was wrong. the quotient group seems like something more trickier.

Could anyone show me to what other (more familiar group) is the one isomorphic to?

Solution 1:

This is an abelian group freely generated by two elements $a,b$ which satisfy $4a-6b=0$. Consider $a'=a-b$ and $b'=2a-3b$. Then $2b'=0$. We have $2a'-b'=b$ and $3a'-b'=a$. Thus we have found new generators. The old relation $4a=6b$ becomes $4(3a'-b')=6(2a'-b')$, which is equivalent to $2b'=0$.

Thus, we have the abelian group freely generated by two elements $a',b'$ such that $2b'=0$. This is the group $\mathbb{Z} ~\oplus ~\mathbb{Z}/2$.

More generally, one can show $\mathbb{Z}^2 / \langle (n,m) \rangle \cong \mathbb{Z} ~\oplus ~\mathbb{Z}/\langle \mathrm{gcd}(n,m) \rangle$ if $(n,m) \neq 0$. In general, the Smith normal form provides an algorithm how to decompose a finitely generated abelian group into cyclic groups.