Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
show 2 more comments

1 Answer 1

up vote 4 down vote accepted

$G$ 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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.