1
$\begingroup$

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!

Help!

  • 0
    What?? Let $G$ be the Klein four group, then $\left[ G,\varphi \right]$ is a group of order 2...2012-08-22
  • 0
    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.2012-08-22
  • 0
    your answer for finite groups was good. Although it didn't place before, my subject is really for G infinite.Sorry!2012-08-22
  • 0
    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 $\mathbb{Z}_2$.2012-08-22
  • 0
    What does mean $\varphi$ let be identity on the Z part?2012-08-22
  • 0
    It means that $\phi$ stabilizes $\mathbb{Z}$, so for $(a,b) \in V\times \mathbb{Z}$, $(a,b)^\phi=(a^\phi,b)$.2012-08-23

0 Answers 0