If $H \subseteq xHx^{-1} \, \forall x \in G $ then we have $H=xHx^{-1} \, \forall x \in G$.

Question

Let $G$ be a group and $H \leq G$.

I want to prove that if

$$H \subseteq xHx^{-1} \, \forall x \in G $$

then we have $$H=xHx^{-1} \, \forall x \in G$$

What I've done:

Let $x \in G$. This implies $x^{-1} \in G$.

So we have $H \subseteq xHx^{-1} \implies Hx \subseteq xH $, and by replacing $x$ with $x^{-1}$ we get

$$H \subseteq x^{-1}Hx \implies xH \subseteq Hx.$$

So we get $xH = Hx \implies xHx^{-1} = H \, \forall x \in G$.

Is my proof correct?

Answer 1

It would be a little cleaner to simply replace $x$ with $x^{-1}$ directly in the initial containment: if $H \subset xHx^{-1}$ for all $x$ then $H \subset x^{-1}Hx$ for all $x$, and the second containment implies $xHx^{-1} \subset H$. Thus $H = xHx^{-1}$ for all $x$.

Note. If $H \subset xHx^{-1}$ for a specific $x$ rather than all $x$, then the previous reasoning breaks down and in fact it need be true that $H = xHx^{-1}$ when $H$ is infinite (of course it would follow if $H$ is finite). See here for an example with $xHx^{-1}$ a proper subset of $H$.