Suppose $G$ is an infinite group and $H$ is an infinite subgroup of $G$.
Let $g\in G$.
Suppose $\forall h\in H\ \exists h'\in H$ such that $gh=h'g$.
Can we conclude that $gH=Hg$?
What if $G$ and $H$ are of finite orders?
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2This corresponds to the definition of a normal subgroup. So, yes $gH=Hg$. – Bernard Nov 20 '16 at 23:59
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This occurs in one of the proofs in my textbook, but they only conclude that $gH\subseteq Hg$. – Siddhartha Nov 21 '16 at 00:05
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1@Bernard Not if $g$ is a fixed element. – anon Nov 21 '16 at 00:14
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@arctic term: Oh! I misread. Thought it was for all $g$. – Bernard Nov 21 '16 at 00:37
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If $g$ is a fixed element, all we can conclude is that $gHg^{-1}\subseteq H$. This does not necessarily imply $xHx^{-1}\subseteq H$ for other elements $x\in G$. Nor does it even mean $gHg^{-1}=H$ since it's possible for the conjugate $gHg^{-1}$ to be a proper subset of $H$. There are some examples here as well as numerous other examples in links. My favorite is letting $\alpha$ be the automorphism of $\mathbb{Q}$ defined by the formula $\alpha(x)=\frac{1}{2}x$, forming the semidirect product $\mathbb{Q}\rtimes\langle \alpha\rangle$, then considering the strict containment $\langle 2\rangle \subset \alpha\langle 2\rangle\alpha^{-1}=\langle 1\rangle$.
The conjugation map $x\mapsto gxg^{-1}$ is bijective, so $|gHg^{-1}|=|H|$, which means if $H$ is finite then the containment $gHg^{-1}\subseteq H$ implies $gHg^{-1}=H$.
If $gHg^{-1}\subseteq H$ for all $g\in G$, then $gHg^{-1}=H$ for all $g\in G$. To see this, replace $g$ with $g^{-1}$ then notice $g^{-1}H(g^{-1})^{-1}\subseteq H$ is equivalent to $H\subseteq gHg^{-1}$, so both containments mean equality.
