Say we have a subgroup H of $\Bbb Z \times \Bbb Z$, with $H \cong \Bbb Z$. Lets choose $H:=\{(n,2n)|n \in \Bbb Z\}$ as an example. I'm confused about how to prove the isomorphism.
I know that a proof exists to show that there is a bijection between $\Bbb Z \times \Bbb Z$ and $\Bbb Z$ (I think it was cantor but that might have just been between $(0,1)$ and $(0,1)^p$), So I'm not overly concerned with this part.As if there is a bijection there , there must be one between a subset of $\Bbb Z \times \Bbb Z$ and $\Bbb Z $ .
But my real question is how can I show there is a homomorphism ( the direct product confuses me a bit).
Would it be something like , define the map $\phi : H \rightarrow \Bbb Z$ s.t. $\phi((n,2m))=n$.
Then $\phi((n,2n)+(m,2m))= n+m=\phi(n,2n)+\phi(m,2m)$
So it is also a homomorphism.
P.S. In a proof to show this would you need to write in full the proof that a bijection exists, or can you just say it is well known that a bijection exists ?
If you can choose the subgroup, put $H=\mathbb{Z}\times\{0\}$ and the trivial map $a\mapsto (a,0)$ is an isomorphism.
In your example, $H=\{(n,2n):n\in\mathbb{Z}\}$ and the map $a\mapsto (a,2a)$ has the inverse map $(a,2a)\mapsto a$, so is an isomorphism.