Let $A$ be the additive group $\mathbb{Z}\oplus\mathbb{Z}$ and let $B$ be the subgroup $\{(5m+7n,2m+4n)\mid m,n\in\mathbb{Z}\}$. Show that $A/B$ is cyclic and determine its order.
I'm absolutely stuck here. Any push in the right direction would be appreciated!
Clearly $\mathbb Z=\{5m+7n|m,n\in \mathbb Z\}$, this means that every coset $(x,y)+B$ contains elements of the form $(0,z)$.
We can thus only work with elements of the form $(0,z)+B$. When are two of these equal?
We have $(0,z_1)+A=(0,z_2)+A\iff (0,z_1-z_2)\in A$ if and only if there is $m$ and $n$ with $5m+7n=0$ and $2m+4n=z_1-z_2$, we have $n=-\frac{5m}{7}$ so $m=7l$ and $n=-5l$, we must now have $14l-20l=z_1-z_2$ which has solution if and only if $z_1-z_2$ is a multiple of $6$.
So the group has order $6$, and it is clearly cyclic, as the cosets can be represented by $(0,0),(0,1),(0,2),\dots,(0,5)$.