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Math Help - [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 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.
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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 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.
    Last edited by NonCommAlg; September 6th 2008 at 11:42 PM. Reason: Added the Remark!
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