An example of a group, a subgroup and an element, satisfying a given condition. Is there a group $G$, a subgroup $H$ and an element $x$, such that $xHx^{-1} \subset H$, but  $xHx^{-1} \neq H$?
Thanks in advance.
 A: Yes, there is such a group, subgroup, and element.
Consider the symmetric group of the integers $\text{Sym}(\mathbb{Z})$. You can think of this as the group of permutations on the integers. Let $H$ be the set of permutations that fix the negative integers. Let $x$ be the permutation satisfying $x(a) = a-1$.
Let's consider $x^{-1}Hx$. First $x$ shifts all the integers down, any element in $H$ fixes the integers in all negative slots, and $x^{-1}$ shifts the integers back up. This means that the integers in slots less than $1$ are all fixed.
Since the group of permutations that fix integers less than $1$ is a subgroup of the permutations that fix negative integers, we have that $x^{-1}Hx \subset H$. But $x^{-1}Hx$ doesn't contain any permutation that shifts the $0$'th entry, so $x^{-1}Hx \neq H$.
Note that I pivotally required an infinite nonabelian group. Any abelian group only has normal subgroups. If the group is finite, then $x^{-1}Hx$ will have the same size as $H$, and so containment means equality.
