Q:

For $H=\{(x,y)\in\mathbb{Z}:3x+4y\equiv 0 \mod 5\}$, find a group isomorphism $\mathbb{Z}^2\to H$

I'm working through group theory problems and I'm trying to do the following exercise: 

For $H=\{(x,y)\in\mathbb{Z}:3x+4y\equiv 0 \mod 5\}$, find a group isomorphism $\mathbb{Z}^2\to H$ and show $[\mathbb{Z}^2:H]=5.$

I tried defining the group isomorphism in $5$ cases, letting the value of $f(x,y)\in H\subset\mathbb{Z}^2$ depend on the value of $y$ modolu $5$, but it didn't work out since the map wasn't even a homomorphism. Also, it is really simple to find an injective homomorpism by $(x,y)\mapsto(5x,5y)$, but that does not really help either, of course. I thought about choosing a map that fixes the second element, and by using an enumeration of all solutions of $3x+4y\equiv 0 \mod 5$ for each value of $y$, mapping $k\geq0$ to the $(k+1)$'th positive solution for $x$ and similarly for $k<0$. However, this also does not seem to give rise to a homomorphism, although it seems (to me) a well-defined bijection.
I think I'm missing some clever trick I can use but I don't see it. Can anyone give me a hint for the homomorphism? I think after that the second part (showing that $[\mathbb{Z}^2:H]=5$ is not too difficult therafter), so my question is mainly about the first part. Any help is much appreciated!

A:

Some ideas for you:
Define, for example,
$$f(1,0):=(2,1)\;,\;\;f(0,1):=(3,-1)$$
and extend the definition to $\;\Bbb Z^2\;$ by $\;\Bbb Z\,-$ linearity ( which is simply to apply the universal property of the free abelian group $\;\Bbb Z^2\;$ on the free generators $\;(1,0),\,(0,1)\;$). Observe that you get for free that $\;f\;$ is a group homomorphism. Now just show it is onto and its kernel has actually index $\;5\;$.

