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