Are all index 2 subgroups of $(\mathbb{Z} /2\mathbb{Z})^n$ isomorphic to $(\mathbb{Z} /2\mathbb{Z})^{n-1}$

Question

In particular, consider the homomorphism from $(\mathbb{Z} /2\mathbb{Z})^n \to {\pm1}$ sending $\{ \epsilon_i \}^n$ to $\prod \epsilon_i$ where $\epsilon_i = \pm1$. The kernal of this homomorphism is an index 2 subgroup, call it G. If it is of the form $(\mathbb{Z} /2\mathbb{Z})^{n-1}$, then what would the isomorphism look like? If it is not of this form, what is the group structure of G?

Answer 1

Look at $(\mathbb{Z} /2\mathbb{Z})^n$ as a vector space over $\mathbb{Z} /2\mathbb{Z}$.

Then every subgroup of index $2$ is a subspace of dimension $n-1$ and so is isomorphic to $(\mathbb{Z} /2\mathbb{Z})^{n-1}$.

Answer 2

If you take a step back, then your group can be characterized as being a finite group $G$ with the property that $g^2=1$ for all $g \in G$. This property is inherited by subgroups and quotients.