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

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.

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Yes, it's possible. I gave an example here. – Arturo Magidin Feb 10 '12 at 17:37
And it's not a stupid question! – Derek Holt Feb 11 '12 at 5:51

## marked as duplicate by Arturo Magidin, t.b., sdcvvc, William, wentaway Sep 2 '12 at 21:48

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.

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And I gave the exact same example here... (-: – Arturo Magidin Feb 10 '12 at 17:42

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 $H$.

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 I like this example, thanks. – Sasha Feb 11 '12 at 8:10