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Thread: [SOLVED] Image of a subgroup under a conjugation

  1. #1
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    [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,
    Thanks in advance,
    Laurent.
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  2. #2
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    Quote Originally Posted by Laurent View Post
    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,
    Thanks in advance,
    Laurent.
    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:

    $\displaystyle gH_kg^{-1}=\left\{\begin{pmatrix}1 & 2kn \\ 0 & 1 \end{pmatrix}:\;n\in\mathbb{Z}\right\} \subset H_k.$
    Last edited by NonCommAlg; Sep 6th 2008 at 10:42 PM. Reason: Added the Remark!
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