Group with exactly two subgroups of index 2

I am looking for a group $G$ such that $G$ has exactly two subgroups of index 2.

I have searched by GAP but I couldn't find it.

Solution 1:

Such a group does not exist. Suppose the group $G$ has subgroups $N \ne M$ of index two. Clearly they are both normal, and $M N = G$. Then $$ \lvert G : M \cap N \rvert = \lvert G : M \rvert \cdot \lvert M : M \cap N \rvert = \lvert G : M \rvert \cdot \lvert M N : N \rvert = \lvert G : M \rvert \cdot \lvert G : N \rvert = 4. $$ So $G / M \cap N$ is a group of order $4$, which is isomorphic to the Klein four-group $V$, as $a^{2} \in M \cap N$ for each $a \in G$.

Since $V$ has three subgroups of index $2$, $G$ has at least three such subgroups, by the correspondence theorem.

Solution 2:

There is a very nice result proved by R.R. Crawford and K.D. Wallace, On the number of subgroups of index 2 - an application of Goursat's Theorem for groups, Math. Magazine, 48, no. 3 (1975). Let $I_2(G)=\#\{H \lt G: |G:H|=2\}$.

Theorem Let $n$ be a non-negative integer. Then there exists a group $G$ with $I_2(G)=n$ if and only if $n=2^k-1$ for some non-negative integer $k$.

The theorem can be generalized as follows. Let $p$ be a prime and $N_p(G)=\#\{H \lhd G: |G:H|=p\}$.

Theorem Let $n$ be a non-negative integer. Then there exists a group $G$ with $N_p(G)=n$ if and only if $n=\frac{p^k-1}{p-1}$ for some non-negative integer $k$.