# Does existence of Index 2 s.g. imply existence of involution element?

I'm studying group theory (intro course) at the moment. Recently I made the error assuming that we can always construct an isomorphism between a quotient group and a subgroup. I've learned that this is false with Q8/Z(Q8) being the desired counterexample.

My question now is, is there a similar counter-example for quotient groups formed with Index 2 normal s.groups? That is, can we disprove (or prove) the proposition that the existence of an Index 2 normal s.group implies an existence of an involution (or Z/2 s.g)?

Regards,

Andrej1122

• Are you asking about finite groups? An infinite cyclic group has a subgroup index two but no element of order two.
– bof
Oct 2 '14 at 23:56
• Thanks. I'm asking in general, but any special cases would be greatly appreciated. Oct 3 '14 at 0:02

In the finite case, suppose $G$ is a group and $H$ is an index $2$ subgroup. Then $|G|$ is even, so by Cauchy's theorem it has an element of order $2$, so it has a subgroup of order $2$.
This holds for any prime $p$, by the same argument.
As @bof pointed out, it is not necessarily true if $G$ is infinite.