Example of a subgroup for normality Give an example of a group $G$ and a subgroup $H$ of $G$, such that for  some $g\in G,\     g^{-1}Hg\subset H$ i.e. $g^{-1}Hg$  is properly contained in $H$.
 A: Let $G$ be the set of affine transformations of $\mathbb R$. That is, $G=\{x\mapsto a x + b: a,b\in{\mathbb R}, a\neq 0\}$. Let $H\subset G$ be the subgroup of integer shifts, $H = \{x\mapsto x+k:k\in {\mathbb Z}\}$, and $g(x) = x/2$. Consider $h\in H$. Let $h(x) = x + k$. Then
$$g^{-1}hg(x) = 2 \cdot (x/2 + k) = x + 2k.$$
That is, $g^{-1}hg$ lies in the subgroup of shifts by an even integer number, $H_2 = \{x\mapsto x+2k:k\in {\mathbb Z}\}$, which is properly contained in $H$.
A: Clearly an example of a subgroup strictly containing a conjugate of it requires an infinite group. The following example is rather large, but easy to understand; maybe there are more elementary examples.
Let $\Gamma$ be the oriented graph on $\mathbf Z\times\mathbf N$ where $(i,j)$ has a single outgoing edge to $(i+1,\lfloor\frac j2\rfloor)$ (so every point has $2$ incoming edges). Let $G$ be the group of automorphisms of $\Gamma$, $H$ the subgroup stabilising $(1,0)$, and $g:((i,j)\mapsto(i-1,j))\in G$. Then $gHg^{-1}$ is the stabiliser of the point $(0,0)$, which contains $H$ since $(1,0)$ is the unique point reachable by an outgoing edge from $(0,0)$. But $G$ contains elements that fix $(1,0)$ but interchange $(0,0)$ with $(0,1)$ (left as exercise), so $H$ strictly contains $gHg^{-1}$.
