Let $G=\{f:\mathbb N\to \mathbb Z\}$ be a commutative group with the following operation for $f,g\in G$ and $n\in\mathbb N$:
$$(f+g)(n):=f(n)+g(n).$$
I've already showed that the set $H=\{h:\in G:\forall n\in\mathbb N,h(n)\in2\mathbb Z\}$ is a subgroup of $G$.
The second request of the exercise asks me to show that $|G:H|$ is infinite.
This part of the problem seemed to be quite easy but I can't find a consistent way to prove this.
I've showed that $H\trianglelefteq G$, since $\forall n\in N$ the conjugated of a generic $h\in H$ with an element of $f\in G$ is such that
$$(\bar f+h+f)(n)=(-f+h+f)(n)=-f(n)+h(n)+f(n)\in H.$$
I was trying to find a surjective homomorphism $\varphi: G\to X$ such that $H$ is the Kernel of this homomorphism in order to apply the first isomorphism theorem with $G/H\cong\varphi (G)\cong X$, where X is an infinite group, but I can't find it. Using the corespondence theorem on $G/H$,supposing that $|G:H|$ is finite, also seems to be inconclusive.
I really thank you for the support.
Following your idea, we can consider the map $\varphi \colon G \rightarrow (\mathbb{Z}/2 \mathbb{Z})^{\mathbb{N}}$ defined by $$\varphi(f) = \left( f(n) \bmod{2} \right)_{n \in \mathbb{N}} \, \text{.}$$ The map $\varphi$ is a surjective group homomorphism whose kernel is $H$, so it induces an isomorphism from $G/H$ to the group $(\mathbb{Z}/2 \mathbb{Z})^{\mathbb{N}}$, which is infinite.