# [SOLVED] Image of a subgroup under a conjugation

• Sep 6th 2008, 05:31 PM
Laurent
[SOLVED] Image of a subgroup under a conjugation
While dealing with a question on the forum, I came up with the following problem:

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

The answer (to the last question) is "no" if $\displaystyle H$ is finite. Or if $\displaystyle H$ 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,
Laurent.
• Sep 6th 2008, 07:17 PM
NonCommAlg
Quote:

Originally Posted by Laurent
While dealing with a question on the forum, I came up with the following problem:

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

The answer (to the last question) is "no" if $\displaystyle H$ is finite. Or if $\displaystyle H$ 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,
here's an example: let $\displaystyle H=\left\{\begin{pmatrix}1 & n \\ 0 & 1 \end{pmatrix}:\;n\in\mathbb{Z}\right\}$. then $\displaystyle H$ is a subgroup of $\displaystyle G=GL(2, \mathbb{Q}).$ let $\displaystyle g=\begin{pmatrix}2 & 0 \\ 0 & 1 \end{pmatrix}.$ then: $\displaystyle gHg^{-1}=\left\{\begin{pmatrix}1 & 2n \\ 0 & 1 \end{pmatrix}:\;n\in\mathbb{Z}\right\} \subset H.$
Remark: in general for a fixed natural number $\displaystyle k$ let $\displaystyle H_k=\left\{\begin{pmatrix}1 & kn \\ 0 & 1 \end{pmatrix}:\;n\in\mathbb{Z}\right\}$. then $\displaystyle H_k$ is a subgroup of $\displaystyle G=GL(2, \mathbb{Q}).$ let $\displaystyle g=\begin{pmatrix}2 & 0 \\ 0 & 1 \end{pmatrix}.$ then:
$\displaystyle gH_kg^{-1}=\left\{\begin{pmatrix}1 & 2kn \\ 0 & 1 \end{pmatrix}:\;n\in\mathbb{Z}\right\} \subset H_k.$