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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).

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Clearly we at least need $n$ to be even. –  Chris Eagle Apr 27 '12 at 16:55

1 Answer 1

up vote 2 down vote accepted

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.

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I'm trying to understand why does n have to be even, I got to the point that any subgroup 'have to be a subgroup in every coordinate' but I can't explain to myself why if the subgroup is of index 2 then in every coordinate we have a subgroup of index 2... –  Belgi Apr 27 '12 at 16:58
    
@Belgi: What point? It is not true that a subgroup "has to be a subgroup in every coordinate"! Just think about the Klein $4$-group, $\mathbb{Z}_2\times\mathbb{Z}_2$. The subgroup $\{(0,0),(1,1)\}$ is not a "subgroup in every coordinate." However, having a subgroup in every coordinate does yield a subgroup of the product. And note that $H$ is not a subgroup of index $2$! Instead, I use $H$ to get to a group which we know has subgroups of index $2$. If $K$ is a subgroup of index $2$ **in $G/H$**, then there exists a subgroup $\mathcal{K}$ of $G$, (cont) –  Arturo Magidin Apr 27 '12 at 17:01
    
@Belgi: such that $H\subseteq\mathcal{K}$, and $\mathcal{K}/H = K$. By the Isomorphism Theorems, $G/\mathcal{K} \cong (G/H)/(\mathcal{K}/H) = (G/H)/K$, and since $(G/H)/K$ has two elements, that means that $[G:\mathcal{K}]=2$. Nowhere did I claim, or even imply, that $H$ itself had index $2$. I'm using the isomorphism theorems: There is a one-to-one, inclusion preserving, index-preserving correspondence between subgroups of $G/H$ and subgroups of $G$ that contain $H$. –  Arturo Magidin Apr 27 '12 at 17:03
    
Thank you for the correction, I also don't understand the claim "all elements of G have odd order" - If I were to deal with a finite group I could of used lagrange theorem, but with an infinite group I'm a bit lost... –  Belgi Apr 27 '12 at 17:16
    
@Belgi: If $g\in G$, then $g=(a_1,\ldots,a_k,\ldots)$, with $a_i\in \mathbb{Z}_n$ for all $i$. Therefore, $ng = (na_1,\ldots,na_k,\ldots)$, and since $a_i\in\mathbb{Z}_n$, then $na_i=0$ for all $i$. So $ng = (0,0,\ldots,0)$, hence the order of $g$ divides $n$, hence the order of $g$ is odd if $n$ is odd. –  Arturo Magidin Apr 27 '12 at 17:18

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