# All subgroups of index 2

Can we construct examples of non-abelian groups G (finite or infinite) such that for all of it's non-trivial subgroups has index 2 in G?.

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Here is another way to prove that in the finite case $G$ has to be abelian: apparently $G$ has even order, whence a Sylow 2-subgroup $S$ is non-trivial. Assume $S \lt G$. Then by the condition it has index $2$, which is impossible, since for the Sylow group $S$, index$[G:S]$ is odd. Hence $G=S$, so $G$ is a 2-group. But then its center $Z(G)$ is non-trivial and has index $2$. So $G/Z(G)$ is cyclic and that implies $G$ is abelian. Further, it is then easy to see that $G \cong C_4$ or $C_2 \times C_2$, the Klein 4-group.

Observation: if $G$ is infinite non-abelian with every non-trivial subgroup having index $2$, then the only subgroups are the trivial ones, $G$ and $\{1\}$. For suppose $H$ is a subgroup, with $H \gt 1$, so index$[G:H]=2$. Pick an $x \in H-\{1\}$, then $\langle x \rangle \subseteq H$, and also index$[G:\langle x \rangle]=2$. Hence $H=\langle x \rangle$. Since $G$ is infinite, $\langle x^2 \rangle$ cannot be trivial (otherwise $x$ would have finite order $2$ and $G$ would be finite). So $\langle x \rangle=\langle x^2 \rangle$. Hence $x=x^{2n}$ for some integer $n$, and this implies that $x$ has finite order after all, a contradiction. We conclude that $G=H$.
So $G$ has only two subgroups and this means $G \cong \mathbb{Z}$, which is abelian, a contradiction. So the infinite case is refuted.

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$\mathbb{Z}$ has infinitely many subgroups, but the point still stands. – Jack Schmidt Feb 26 '14 at 3:56
Jack, you are right! Thanks. Double contradiction. – Nicky Hekster Feb 26 '14 at 4:10

Not of finite order. Suppose $G$ has subgroup of index 2. Then order of $G$ is even, as it's a union of two cosets of such subgroup. By Cauchy's Theorem, $G$ has member of order two. If $|G|>4$, index of the subgroup generated by this element is greater than 2. If $G$ has no subgroups at all, it is cyclic (of prime order), hence abelian.