Subgroups of $\mathbb{Z}_2^n$ of order $2^{n-1}$

44 Views Asked by At

What are the subgroups of $\mathbb{Z}_2^n$ of size $2^{n-1}$?

I'm fairly convinced that it will be subgroups of the form $\mathbb{Z}_2 \times \dots \times \{0\} \times \dots \times \mathbb{Z}_2$, but I can't seem to know how to prove it.

edit: my guess is wrong. take $n=2$ and $\{(1 1)(00)\} \leq \mathbb{Z}_2^2$.

1

There are 1 best solutions below

0
On BEST ANSWER

$\;V:=\left(\Bbb Z_2\right)^n\;$ is an $\;n\,-$ dimensional vector space over the field $\;\Bbb F_2\cong\Bbb Z_2\;$ , and there's a $1$-$1$ correspondence between the subgroups of $\;V\;$ and the subspaces of $\;V\;$ . Since a group $\;H\le V\;$ has $\;2^{n-1}\;$ elements iff $\;\dim H=n-1\;\iff\; H\;$ is a hyperplane of $\;V\;$ , i.e. a maximal proper subspace of $\;V\;$.

We know that hyperplanes are in fact the kernel of non-zero linear functionals $\;\phi:V\to\Bbb F_2\;$ , so if you know how many of these are we're done...or you can also argue that the number of subspaces of dimension $\;n-1\;$ in $\;V\;$ is equal to the number of $\;1\,-$ dimensional subspaces (why?), and these are easier to calculate...