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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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    Sorry, i tried :\ (+1)2012-04-26

1 Answers 1

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$G$ 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.

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    This is a great answer, but I believe it doesn't hold for $Z_4$ for example, is the claim still true ?2012-04-26