Conjugate subgroup strictly contained in the initial subgroup? [duplicate]

Question

Possible Duplicate:
The set of all $x$ such that $xHx^{-1}\subseteq H$ is a subgroup, when $H\leq G$

Probably a very stupid question:

Let $G$ be a group, $H\subset G$ a subgroup, $a\in G$ an element. Is it possible that $aHa^{-1} \subset H$, but $aHa^{-1} \neq H$? If $H$ has finite index or finite order, this is not possible.

Answer 1

Consider the group of matrices $$G=\left\{ \begin{bmatrix} x & y \\ 0 & 1 \end{bmatrix} : x,y \in \mathbb{Q} \right\} = \operatorname{AGL}(1,\mathbb{Q}) $$ and its subgroup $$H=\left\{ \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} : y \in \mathbb{Z} \right\} \cong \mathbb{Z}$$ and of course the single element $$a=\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$$ A direct calculation gives $$aHa^{-1} = \left\{ \begin{bmatrix} 1 & 2y \\ 0 & 1 \end{bmatrix} : y \in \mathbb{Z} \right\} < H$$ is a proper subgroup of H.

Similar issues showed up in this question.

Answer 2

First of all, the group $G$ has to be non-abelian. Otherwise any subgroup is normal and $a^{-1}Ha = H$ for each $a \in G$. You also need the subgroup $H$ (and thus $G$) to be infinite as you mentioned.

I'll give a different example. Let $G = S_{\mathbb{Z}}$, the group of bijections from $\mathbb{Z}$ to $\mathbb{Z}$.

For the subgroup, let $H = \{f \in G \mid f(x) = x \text{ for each } x \leq 0 \}$. Define the map $\sigma \in G$ by $\sigma(x) = x + 1$ for every $x \in \mathbb{Z}$.

It is not difficult to show that $\sigma H\sigma^{-1} \subseteq H$. Since $\sigma f \sigma^{-1}(1) = \sigma(f(0)) = \sigma(0) = 1$ for any $f \in H$, we notice that $\sigma H \sigma^{-1}$ contains only maps that fix $1$. Thus $\sigma H \sigma^{-1}$ cannot be all of $H$.