If $G$ is a group, $H$ is a subgroup of $G$ and $g\in G$, is it possible that $gHg^{-1} \subset H$? [duplicate]

If $G$ is a group, $H$ is a subgroup of $G$ and $g\in G$, is it possible that $gHg^{-1} \subset H$ ?

This means, $gHg^{-1}$ is a proper subgroup of $H$. We know that $H \cong gHg^{-1}$, so if $H$ is finite then we have a contradiction since the isomorphism between the two subgroups implies that they have the same order so $gHg^{-1}$ can't be proper subgroup of $H$.

So, what if $H$ is infinite is there an example for such $G , H , g$ ?

Edit: I suppose that $H$ has a subgroup $N$ such that $N$ is a normal subgroup of $G$.

Solution 1:

Yes. Define $\bf Q$ under addition. Let $\alpha:x\mapsto nx$ be an automorphism for some integer $n>1$. Consider the subgroup $Z$ generated by $(1,0)\in {\bf Q}\rtimes\langle \alpha\rangle$. Then $\alpha Z\alpha^{-1}=nZ\subsetneq Z$.

Solution 2:

Let $\mathbb{F}_2 = \langle a,b \mid \ \rangle$ be the free group of rank two. It is known that the subgroup $F_{\infty}$ generated by $S= \{b^nab^{-n} \mid n \geq 0 \}$ is free over $S$. Then $bF_{\infty}b^{-1}$ is freely generated by $bSb^{-1}= \{b^n a b^{-n} \mid n \geq 1\}$, hence $bF_{\infty}b^{-1} \subsetneq F_{\infty}$.

(Otherwise, $a$ can be written over $bSb^{-1}$, which is impossible since $a \in F_{\infty}$ and $F_{\infty}$ is free over $S$.)