Does there exist $G$ such that for a subgroup $H$ of $G$ , $gHg^{-1}$ is proper in $H$ for some $g\in G$ ?
It is clear that $H,G$ must be infinite. I look for examples in matrice groups and not found yet.
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Yes. There should be several questions about this with answers. – zibadawa timmy Oct 29 '14 at 18:34
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1@zibadawatimmy: Can you give link please ? – mesel Oct 29 '14 at 18:34
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1This was discussed for example here. I've also seen more natural exaples ($G=S(\mathbb{Z})$, $H$ subgroup fixing all negative elements, and $g=(k \mapsto k+1)$, if I remember correctly) but cannot find it at the moment. – Pavel Čoupek Oct 29 '14 at 18:45
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1Here it is. – Pavel Čoupek Oct 29 '14 at 18:52
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An older question, with more answers: http://math.stackexchange.com/questions/107862/conjugate-subgroup-strictly-contained-in-the-initial-subgroup – user1729 Oct 29 '14 at 20:01
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Take $G$ the group of affine transformations of $\mathbb{Q}$ , $G= \{(a,b) \ | \ a \in \mathbb{Q}^{\times}, b \in \mathbb{Q}\}$ with $ (a,b)$ acting as $t \mapsto a t + b$. Take $g = (2,0)$ and $H = \{1\} \times \mathbb{Z}$. We have $g H g^{-1} = \{1\} \times 2\mathbb{Z}$.
As groups of matrices let $G = $ matrices of form $ \left( \begin{array}{cc} a & b\\ 0 & 1\\ \end{array} \right)$ with $a \in \mathbb{Q}^{\times}$, $b \in \mathbb{Q}$ and $H $ matrices of form $ \left( \begin{array}{cc} 1 & n\\ 0 & 1\\ \end{array} \right)$ with $n \in \mathbb{Z}$
The idea was to use a semidirect product $L \rtimes T$ where $T$ is a group of automorphisms of $L$ that contains a transform taking a subgroup $H$ of $L$ to a proper subgroup of $H$.
orangeskid
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