[SOLVED] Image of a subgroup under a conjugation

**Laurent** · September 6th 2008, 05:31 PM

While dealing with a question on the forum, I came up with the following problem:

If $H$ is a subgroup of a group $G$ , and $g\in G$ , does $gHg^{-1}\subset H$ imply $gHg^{-1}=H$ ? In other words, is it possible to have $gHg^{-1}\subsetneq H$ ?

The answer (to the last question) is "no" if $H$ is finite. Or if $H$ is a normal subgroup. But what else?

I think this is either easy or false, but I can find neither a proof nor a counterexample...

I hope some people can help,
Thanks in advance,
Laurent.

**NonCommAlg** · September 6th 2008, 07:17 PM

Originally Posted by Laurent

While dealing with a question on the forum, I came up with the following problem:

If $H$ is a subgroup of a group $G$ , and $g\in G$ , does $gHg^{-1}\subset H$ imply $gHg^{-1}=H$ ? In other words, is it possible to have $gHg^{-1}\subsetneq H$ ?

The answer (to the last question) is "no" if $H$ is finite. Or if $H$ is a normal subgroup. But what else?

I think this is either easy or false, but I can find neither a proof nor a counterexample...

I hope some people can help,
Thanks in advance,
Laurent.

here's an example: let $H=\left\{\begin{pmatrix}1 & n \\ 0 & 1 \end{pmatrix}:\;n\in\mathbb{Z}\right\}$ . then $H$ is a subgroup of $G=GL(2, \mathbb{Q}).$ let $g=\begin{pmatrix}2 & 0 \\ 0 & 1 \end{pmatrix}.$ then: $gHg^{-1}=\left\{\begin{pmatrix}1 & 2n \\ 0 & 1 \end{pmatrix}:\;n\in\mathbb{Z}\right\} \subset H.$

Remark: in general for a fixed natural number $k$ let $H_k=\left\{\begin{pmatrix}1 & kn \\ 0 & 1 \end{pmatrix}:\;n\in\mathbb{Z}\right\}$ . then $H_k$ is a subgroup of $G=GL(2, \mathbb{Q}).$ let $g=\begin{pmatrix}2 & 0 \\ 0 & 1 \end{pmatrix}.$ then:

$gH_kg^{-1}=\left\{\begin{pmatrix}1 & 2kn \\ 0 & 1 \end{pmatrix}:\;n\in\mathbb{Z}\right\} \subset H_k.$

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