Finding subgroups of index 2 of $G = \prod\limits_{i=1}^\infty \mathbb{Z}_n$ I looked at this question and its answer. The answer uses the fact that every vector space has a basis, so there are uncountable subgroups of index 2 if $n=p$ where $p$ is prime.
Are there uncountable subgroups of index 2 if $n$ is not prime ?
The problem looks the same (with minimal change), but the way we found the subgroups is not good for this case (I think).

If $n$ is odd, $G$ has no subgroups of index $2$. Indeed, if $H$ is a subgroup of index dividing $2$, and $g\in G$, then $2g\in H$ (since $G/H$ has order $2$, so $2(g+H) = 0+H$). Since every element of $G$, hence of $H$, has order dividing $n$, and $\gcd(2,n)=1$, then $\langle 2g\rangle = \langle g\rangle$, so $g\in\langle 2g\rangle\subseteq H$, hence $g+H$ is trivial. That is, $G\subseteq H$. So the only subgroup of index dividing $2$ is $G$.
If $n$ is even, then let $H=2\mathbb{Z}_n$, which is of index $2$ in $\mathbb{Z}_n$. Then $G/\prod_{i=1}^{\infty} H \cong \prod_{i=1}^{\infty}(\mathbb{Z}_n/2\mathbb{Z}_n) \cong \prod_{i=1}^{\infty}\mathbb{Z}_2$. 
Note: $\prod_{i=1}^{\infty} H$ is not itself of index $2$ in $G$; in fact, $\prod_{i=1}^{\infty} H$ has infinite index in $G$. We are using $\prod_{i=1}^{\infty}H$ to reduce to a previously solved case.
Since $\prod_{i=1}^{\infty}\mathbb{Z}_2$ has uncountably many subgroups of index $2$ by the previously solved case, by the isomorphism theorems so does $G$.

Below is an answer for the wrong group (I thought $G=\prod_{n=1}^{\infty}\mathbb{Z}_n$; I leave the answer because it is exactly the same idea.)
For each $n$, let $H_n$ be the subgroup of $\mathbb{Z}_n$ given by $2\mathbb{Z}_n$. Note that $H_n=\mathbb{Z}_n$ if $n$ is odd, and $H_n$ is the subgroup of index $2$ in $\mathbb{Z}_n$ if $n$ is even.
Now let $\mathcal{H}=\prod_{n=1}^{\infty}H_n$. Then 
$$G/\mathcal{H}\cong\prod_{n=1}^{\infty}(\mathbb{Z}_n/H_n) = \prod_{n=1}^{\infty}(\mathbb{Z}/2\mathbb{Z}).$$
(In the last isomorphism, the odd-indexed quotients are trivial, the even-indexed quotients are isomorphic to $\mathbb{Z}/2\mathbb{Z}$; then delete all the trivial factors).
Since $G/\mathcal{H} \cong \prod_{n=1}^{\infty}(\mathbb{Z}/2\mathbb{Z})$ has uncountably many subgroups of index $2$, so does $G$ by the isomorphism theorems.
