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Give an example of a group $G$ and a subgroup $H$ of $G$, such that for some $g\in G,\ g^{-1}Hg\subset H$ i.e. $g^{-1}Hg$ is properly contained in $H$.

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Do you know how to accept answers to your questions? I suggest you read up on it, as it's generally considered a good idea. –  Gerry Myerson Sep 8 '12 at 12:42
ok !!Sir...I edited this.... –  ram Sep 8 '12 at 12:46
If you really mean properly contained, that is $g^{-1}H g\subset H$ does not allow the case $g^{-1}H g= H$, then no such example exists as the condition fails for $g=1$. Without "properly", any normal subgroup will do. –  Hagen von Eitzen Sep 8 '12 at 12:48
No such example exists, as $g^{-1}Hg \subseteq H$ implies (by applying $x \mapsto gxg^{-1}$) $H \subseteq gHg^{-1}$ and hence: If $g^{-1}Hg \subset H$ for all $g$, then $H \subset gHg^{-1}$ for all $g$. Impossible. –  martini Sep 8 '12 at 12:49
I think your accept rate would drive trivial any normal subgroup. Seems like you don't like the answers you've received here... –  DonAntonio Sep 8 '12 at 13:09
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marked as duplicate by user1729, Jack Schmidt, rschwieb, Lord_Farin, Julian Kuelshammer Jun 18 '13 at 17:10

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Let $G$ be the set of affine transformations of $\mathbb R$. That is, $G=\{x\mapsto a x + b: a,b\in{\mathbb R}, a\neq 0\}$. Let $H\subset G$ be the subgroup of integer shifts, $H = \{x\mapsto x+k:k\in {\mathbb Z}\}$, and $g(x) = x/2$. Consider $h\in H$. Let $h(x) = x + k$. Then $$g^{-1}hg(x) = 2 \cdot (x/2 + k) = x + 2k.$$ That is, $g^{-1}hg$ lies in the subgroup of shifts by an even integer number, $H_2 = \{x\mapsto x+2k:k\in {\mathbb Z}\}$, which is properly contained in $H$.

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Clearly an example of a subgroup strictly containing a conjugate of it requires an infinite group. The following example is rather large, but easy to understand; maybe there are more elementary examples.

Let $\Gamma$ be the oriented graph on $\mathbf Z\times\mathbf N$ where $(i,j)$ has a single outgoing edge to $(i+1,\lfloor\frac j2\rfloor)$ (so every point has $2$ incoming edges). Let $G$ be the group of automorphisms of $\Gamma$, $H$ the subgroup stabilising $(1,0)$, and $g:((i,j)\mapsto(i-1,j))\in G$. Then $gHg^{-1}$ is the stabiliser of the point $(0,0)$, which contains $H$ since $(1,0)$ is the unique point reachable by an outgoing edge from $(0,0)$. But $G$ contains elements that fix $(1,0)$ but interchange $(0,0)$ with $(0,1)$ (left as exercise), so $H$ strictly contains $gHg^{-1}$.

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