Does $gHg^{-1}\subseteq H$ imply $gHg^{-1}= H$? [duplicate]

Solution 1:

Another example. Let $$ K = \left\{ \frac{a}{2^{n}} : a \in \mathbf{Z}, n \in \mathbf{N} \right\} $$ be the additive subgroup of $\mathbf{Q}$. The map $g : x \mapsto 2 x$ is an automorphism of $K$. Consider the semidirect product $G = K \rtimes \langle g \rangle$. (So that conjugating an element $x$ of $K$ by $g$ in $G$ is the same as taking the value of $g$ on $x$.) Let $H = \left\{ \frac{a}{2} : a \in \mathbf{Z} \right\}$ be a subgroup of $G$. Then $H^{g} = \mathbf{Z} < H$.

PS When I was first exposed to these examples, what struck me is what happens if you look at it from the other end: $\mathbf{Z}^{g^{-1}} = H > \mathbf{Z}$.