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I am curious whether it's possible for the group $G$ such that there is a subgroup $H ≤ G$ and an element $g ∈ G$ such that $gHg^{−1}$ $\subsetneq$ $H$ to exist.

It means there exist some duplication under conjugacy operation on $H$, but is it possible?

Beverlie
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    There can't be any duplication under conjugation, since $gh_1g^{-1}=gh_2g^{-1}$ implies $h_1=h_2$. But for infinite groups, one doesn't need duplication.... – Greg Martin Apr 18 '18 at 06:31
  • @GregMartin I think I have a tendency to think in finite sets usually.. thx – Beverlie Apr 18 '18 at 07:12

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Yes, for infinite groups of course. As an example consider the group of two by two nonsingular matrices over, say $\Bbb Q$. Then $T=\pmatrix{1&1\\0&1}$ generates a cyclic subgroup $H$. But in $G$, $T$ is conjugate to $T^2$, and $T^2$ generates a proper subgroup of $H$.

Angina Seng
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