# Finding subgroups of index 2

Let $G = \prod_{i=1}^\infty \mathbb{Z}_2$ with addition mod 2. I am trying to find subgroups of index 2. I see that taking the entire space and removing all sequences which have a 1 in a certain position gives a subgroup of index 2. For example the set of all sequences $\{(0,\cdot,\cdot,\ldots)\}$ which have a 0 in the first position forms a subgroup. Taking the coset $\{(0,\cdot,\cdot,\ldots)\} + (1,0,0,\ldots)$ gives all other sequences, so there are two cosets.

This gives me infinitely many subgroups of index 2. However, I have read there are actually uncountably many such subgroups. How can I find the others? Thank you!

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This should give you uncountable subgroups. taking out the 1's in a certain position can be done in more than a countable ways – Belgi Apr 26 '12 at 15:43
There are only countably many positions to remove a single 1 from. How can that be done in uncountably many ways? – nullUser Apr 26 '12 at 15:45
What about taking out the 1's in the indexes in the set A, where A can be any subset of $\mathbb{N}$ ? – Belgi Apr 26 '12 at 15:47
@Belgi: That won't give a subgroup of index $2$ in general. – Chris Eagle Apr 26 '12 at 15:50
I considered that but it doesn't work. Suppose I removed the ones from the first two positions. I.e. taking $\{ 0,0, \cdot,\ldots \}$. Looking at the coset $\{ 0,0, \cdot,\ldots \}+(0,1,0,0,\ldots)$ gives all sequences which begin (0,1) which is NOT all the other sequences. Hence there are at least 3 cosets, so the index will be at least 3. – nullUser Apr 26 '12 at 15:50

$G$ is a vector space over the field $\mathbb{Z}_2$. $G$ is uncountable so it must be uncountable-dimensional (in fact it's continuum-dimensional but we don't need that). If $B$ is a basis and $b$ is in $B$, then the span of $B \setminus \{ b \}$ is a subspace of codimension $1$, and hence a subgroup of index $2$. Since $B$ is uncountable, this gives uncountably many such subgroups.
This proof makes use of the axiom of choice in asserting that $G$ as a $\mathbb{Z}_2$-vector space has a basis. I believe this is essential (i.e. that it's consistent with ZF that $G$ has only countably many subgroups of index $2$) but I'm not certain.
Note that any nonzero element is omitted by some (actually infinitely many) index-2 subgroup: for any element $0\neq b\in G$, you can find a basis $B$ that contains $b$, and then apply Chris's answer. – Brett Frankel Apr 26 '12 at 16:35
This is a great answer, but I believe it doesn't hold for $Z_4$ for example, is the claim still true ? – Belgi Apr 26 '12 at 21:45