Is $\mathbb{Z}\times\mathbb{Z}/\langle(6,5),(3,4)\rangle$ finitely generated?

Question

Let $A$ be the quotient of the free abelian group $\mathbb{Z}^2$ by the subgroup generated by $(6,5)$ and $(3,4)$. The question is $A$ finitely generated? And if yes can we decompose it into a direct some of cyclic groups?

My attempt:-
The answer for the first part is yes. $A$ is finitely generated by two elements let's call these $x,y$ hence we will get a system of equations $$ 6x+3y=0$$ $$ 5x+4y=0$$ So basically we have this matrix to deal with it :-

\begin{pmatrix} 6 & 3 \\ 5 & 4 \end{pmatrix} and from here I stopped, any help appreciated.

Answer 1

The Smith Normal Form of your matrix is $\left(\begin{smallmatrix}1&0\\0&9\end{smallmatrix}\right)$ hence $A\cong \mathbb{Z}/9\mathbb{Z}$.

More generally, the SNF will be a block diagonal matrix $\left(\begin{smallmatrix}I&0&0\\0&D&0\\0&0&0\end{smallmatrix}\right)$. Ignore the first block of columns and rows, leaving $\left(\begin{smallmatrix}D&0\\0&0\end{smallmatrix}\right)$. $D$ will be a diagonal matrix $d_1, d_2, \ldots, d_k$, with $d_1|d_2|\cdots|d_k$, and there will be $m$ additional columns. Then your group is isomorphic to $$\mathbb{Z}^m\times (\mathbb{Z}/d_1\mathbb{Z})\times\cdots\times (\mathbb{Z}/d_k\mathbb{Z})$$

Answer 2

Since $(\det A) A^{-1} = \operatorname{adj} A$ has integer entries, we get that $\langle (6,5),(3,4)\rangle$ has index at most $\det A$ and so the quotient is finite of order dividing $\det A = 9$.

Indeed, let $v_1=(6,5)$, $v_2=(3,4)$, $e_1=(1,0)$, and $e_2=(0,1)$. Then $$ \begin{pmatrix} v_1\\ v_2 \end{pmatrix} = \begin{pmatrix} 6 & 5 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} e_1\\ e_2 \end{pmatrix} $$ Writing this as $V=AE$ and letting $B=\operatorname{adj} A$, we get $BV=(\det A) E$, which implies that $9(\mathbb{Z}\times\mathbb{Z}) \subseteq \langle (6,5),(3,4)\rangle$