Is conjugation in infinite groups well behaved?

Asked Jan 10 '17 at 00:43

Active Jan 10 '17 at 00:48

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A question from earlier today made me start to think of this question. Let $G$ be an infinite group with infinite subgroup $H$. Can the condition below ever hold?

\begin{equation*} \exists g\in G \text{ such that } gHg^{-1}\subset H \end{equation*}

where the containment above is strict.

edited Jan 10 '17 at 00:48

asked Jan 10 '17 at 00:43

user404127

1

I take it $gHg^{-1} \subset H$ means that $gHg^{-1}$ is a proper subset of $H$? – pjs36 Jan 10 '17 at 00:47
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Yes, absolutely! – user404127 Jan 10 '17 at 00:47
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Yes, and this has been asked many times before. On the right of my screen, I am seeing a list of related questions, and the very first of these is the identical question to yours. – Derek Holt Jan 10 '17 at 00:51
@DerekHolt Thanks for pointing that out to me; I'm still learning this site! – user404127 Jan 10 '17 at 01:20

Is conjugation in infinite groups well behaved?

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