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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.

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    I take it $gHg^{-1} \subset H$ means that $gHg^{-1}$ is a *proper* subset of $H$? – 2017-01-10
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    Yes, absolutely! – 2017-01-10
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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. – 2017-01-10
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    @DerekHolt Thanks for pointing that out to me; I'm still learning this site! – 2017-01-10

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