Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Set $\varphi \in \operatorname{Aut}\left( G\right) $ with order $2$. Consider the semidirect product $G\rtimes \left\langle \varphi \right\rangle$. ($G$ is infinite)

Let $\left[ G,\varphi \right] =\left\langle g^{-1}g^\varphi ;g\in G\right\rangle \trianglelefteq G.$

My question is: Does $\left[ G,\varphi \right] $ has involutions?

I think doesn't have!


share|cite|improve this question
What?? Let $G$ be the Klein four group, then $\left[ G,\varphi \right]$ is a group of order 2... – user641 Aug 22 '12 at 22:09
I've voted to close. Your original question was easily seen to be false, and after I gave a counterexample, you changed the question. You changed it without thinking it would seem, because my counterexample can easily be extended to cover this new case as well. I am not confident you have given much thought to this question, and without any motivation, I voted to close. – user641 Aug 22 '12 at 22:17
your answer for finite groups was good. Although it didn't place before, my subject is really for G infinite.Sorry! – User2040 Aug 22 '12 at 22:25
Just to follow up on Steve D's comment, you can cross the Klein four group with $\mathbb{Z}$ and just let $\phi$ be the identity on the $\mathbb{Z}$ part. Before you edit again, consider taking $G$ to be an infinite direct sum of copies of $\mathbb{Z}_2$. – MartianInvader Aug 22 '12 at 22:26
What does mean $\varphi$ let be identity on the Z part? – User2040 Aug 22 '12 at 22:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.