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Here is the question: Given a group $G$ with exactly $n$ subgroups of index $2$. I would like to prove that there exists a finite abelian group $H$ that has exactly $n$ subgroups of index $2$.

I was thinking as all subgroups $K$ has index $2$, they are normal so we can perform a quotient group, therefore $g^2\in K.$ Now this looks like a previous exercise I did, using that all $g^2=e$ in additive law means $g+g=e$.

I thin we can look at $\Bbb{Z}/(2)$ this is an abelian subgroup of order $2$ and then there is one-to-one correspondence between subgroups of index two and homomorphisms $G\to \mathbb Z/(2).$

What is the next step?

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Let $N$ be the intersection of the $n$ subgroups $K_1,\ldots, K_n$ of index $2$ in $G$. Being the intersection of normal, $N$ is normal. Let $H=G/N$. You already showed that $g^2=1$ for all $g\in H$ and hence $H$ is abelian. The subgroups of index $2$ in $H$ turn out to be the $K_i/N$.

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  • $\begingroup$ Thanks, I was very close. :) $\endgroup$ – JeSuis Oct 26 '15 at 16:47

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