TITLE: Classify $\mathbb{Z_6} \times \mathbb{ Z_{24}} / \langle(3,2)\rangle$ according to fundamental theorem of finitely generated abelian groups
QUESTION [8 upvotes]: Classify $\mathbb{Z_6} \times \mathbb{ Z_{24}} / \langle(3,2)\rangle$ according to fundamental theorem of finitely generated abelian groups.

The order of $G/H = 12$
So it can be isomorphic to 
$\mathbb{Z_3} \times \mathbb{Z_4}$ or 
$\mathbb{Z_3} \times \mathbb{Z_2} \times \mathbb{Z_2}$
$(0,1)$ has order of 4, 
$(1,0)$ has order of 12,
$(1,1)$ has order of 24
so I'm picking group which has order 4, in this case isomorphic to $\mathbb{Z
_3} \times \mathbb{Z_4}$. 
a) Is this correct and any other better way to do it?
b) If I were to use $3a+2b =0$, how can I find the isomorphic group (not too sure how to do it)?
Thanks for the help folks

REPLY [3 votes]: The group in question is generated by $a,b$ satisfying $6a=0$, $24b=0$, $3a+2b=0$. In matrix form, this is
$$
\pmatrix{ 6 & 0 \\ 0 & 24 \\ 3 & 2 }
$$
Its Smith normal form is
$$
\pmatrix{ 1 & 0 \\ 0 & 12 \\ 0 & 0 }
$$
which proves that the group is $C_{12}$.
Following the Smith reduction (or by hand), we get that the group is generated by $a,c$ with $c=a+b$ and then $a+2c=0$, $12c=0$. Therefore, $c=a+b$ is a generator of the group.