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It is well known that the following three definitions of a normal subgroup are equivalent:

$gNg^{-1}\supseteq N$ for all $g\in G$

$gNg^{-1}\subseteq N$ for all $g\in G$

$gNg^{-1} =N$ for all $g\in G$

However this equality comes from the fact both $g$ and $g^{-1}$ satisfy the requirements.

Can we have a subgroup $H$ such that $gHg^{-1}\subsetneq H$? clearly such a subgroup must be both of finite order and non-normal, which is one of the reasons I can't find a counterexample.

It can also be proven without much difficulty if such a $G,H, g$ existed then $g^{-1}$ would satisfy $g^{-1}Hg\supsetneq H$

Thank you kindly

Regards.

Asinomás
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1 Answers1

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Let $G = \langle a,b \, : \, aba^{-1} = b^{2}\rangle$, and let $H = \left<b\right>$. Then $aHa^{-1} = \langle b^{2}\rangle \subsetneq H$. None of the groups here are finite.

Alastair Litterick
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