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Suppose that groups G, H are such that $H \lhd G$, $H \cong (\mathbb{Z}/2\mathbb{Z})$, and $G / H \cong (\mathbb{Z}/2\mathbb{Z})^n$ for some $n \geq 1$. What then can we say about $G$? I think the only possibilities are that either $G \cong (\mathbb{Z} / 2 \mathbb{Z})^{n+1}$ or $G \cong (\mathbb{Z}/4\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z})^{n-1}$, but I'm not sure how you could prove anything like this.

Rafa
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    Actually, $G$ might not even be abelian. For example take $G=Q$ (the quaternions) and $H={1,-1}$. The quotient is isomorphic to $(\mathbb{Z}/\mathbb{2Z})^2$. – Mark Dec 03 '20 at 14:02

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Writing $C_2$ for $\Bbb Z/2\Bbb Z$ we have a short exact sequence of groups $$ 1\rightarrow C_2\rightarrow G\rightarrow C_2^n\rightarrow 1. $$ So $G$ is a group extension of $C_2^n$ by $C_2$. This is classified up to equivalence by the second cohomology group $H^2(C_2^n,C_2)$.

The isomorphism classes are "easier" to see. Indeed, we may have the (semi)direct product if the sequence splits or some other possibilities if it doesn't split. Note that $G$ need not be abelian. Take for example the quaternion group.

Reference: How to find all possible extensions of a finite group by $C_2$

Some more details are also given here, but with $C_2$ and $C_2^n$ interchanged. Nevertheless it is useful.

Dietrich Burde
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  • Great, thanks for this! – Rafa Dec 03 '20 at 14:20
  • You are welcome! – Dietrich Burde Dec 03 '20 at 14:20
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    I think the group $G$ is a direct product of an elementary abelian group with a group $E$, where $E$ is either cyclic of order $4$, an extraspecial group (of which there are two isomorphism types for each order $2^{2k+1}$), or a group of symplectic type (which is a central product of an extraspecial group with cyclic $4$). – Derek Holt Dec 03 '20 at 14:36