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What is the number of subgroups of index $2$ in $(\mathbb{Z}/2\mathbb{Z})^3$?

I know that a subgroup of index $2$ in $(\mathbb{Z}/2\mathbb{Z})^3$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$ because of the order of its elements, but I cannot say what is the number of this subgroups.

Also, is it possible to generalise this to the number of subgroups of index $2$ of $(\mathbb{Z}/2\mathbb{Z})^n$?

Shaun
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Luigi Traino
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2 Answers2

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If you pick two distinct non-identity elements they are going to generate such a subgroup.

So the answer is $\binom{7}{2}$.

But wait ! Each subgroup can be generated in three ways like this.

Hence the answer is $\binom{7}{2}/3 = 7$

If you want to generalize it to $\mathbb Z_2^n$ you can use linear algebra. Notice that the map that sends each subspace of dimension $n-1$ to its orthogonal complement under the obvious inner product is a bijection between dimension $n-1$ subspaces and dimension $1$ subspaces (to prove bijectivity you can use the $e_i$ vectors, and see here)

Hence there are $2^n-1$ subgroups of index $2$.

Shaun
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Asinomás
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    You can also see the answer for general $p$, where the big space had dimension $n$ and the desired space has dimension $k$. The solution is sort of a generalization of the first solution in my post. There are $\prod\limits_{i=0}^{k-1} (p^n-p^i)$ ways to select a basis for the subspace we want. And each such subspace has $\prod\limits_{i=0}^{k-1} p^k-p^i$ basis. So the final answer is just this quotient. – Asinomás Jul 11 '21 at 17:12
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Any subgroup of index $2$ is normal, so your question is equivalent to "how many surjective homomorphisms are there from $(\mathbb{Z/2Z})^3$ to $\mathbb{Z/2Z}$?". So let's answer that question.

Consider the obvious $3$-element generating set (basis) for $(\mathbb{Z/2Z})^3$. Every homomorphism is defined entirely by where the elements of this set are sent to, which gives $2^3$ homomorphisms (a power of $2$ as each element has two possible images). However, we must exclude the situation where every generator is sent to the identity (as the homomorphism must be surjective), which gives $2^3-1=7$ surjective homomorphisms. Hence, there are $7$ subgroups of index $2$.

Generalising to $(\mathbb{Z/2Z})^n$: bases now have size $n$ which each basis element still has two possible images, hence we get $2^n$ homomorphisms $(\mathbb{Z/2Z})^n\rightarrow \mathbb{Z/2Z}$, so $2^n-1$ surjective homomorphisms/subgroups of index $2$.

Shaun
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user1729
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  • The group is abelian. Every subgroup is normal. – Arturo Magidin Jul 11 '21 at 19:09
  • @Arturo True, although the idea behind the proof works much more generally (e.g. free products of cyclic groups of fixed-prime order rather), so not relying on abelian-ness isn't such a bad thing. – user1729 Jul 11 '21 at 19:17
  • Huh? In the free product of cyclic groups of order $p$, you cannot argue that subgroups of index $p$ are necessarily normal. While it is true that subgroup of index $p$ in a finite group whose order is not divisible by any prime smaller than $p$ is necessarily normal, that argument doesn't work in infinite groups. You are relying on the subgroup being normal, that's fine. The fact that it is index $2$ is irrelevant. – Arturo Magidin Jul 11 '21 at 19:36
  • @Arturo Yes, your right free products don't work! An example which actually works is finite $p$-groups. (All I'm trying to say is that being abelian is a cheap way of guaranteeing normality, and the result you quote about finite groups is much more interesting :-) ) – user1729 Jul 11 '21 at 20:08