Could anyone give me example of the following?
A group $G$, a subgroup $H$, $a$ is a fixed element in $G$, such that $aH$ is a proper subset of $Ha$ and $aH \neq Ha$.
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The fact that this is a duplicate is because if $Ha\subsetneq aH$ then $a^{-1}Ha\subsetneq H$. It is interesting to note that the examples given here are basically identical to the examples in the original question! – user1729 Aug 05 '13 at 14:44
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Let $G = GL_2(\mathbb R)$. $H = \left\{ \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} : a \in \mathbb Z\right\}$ is a subgroup of $G$. Let $a = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$.
$$aH = \left\{ \begin{pmatrix} 2 & 2a \\ 0 & 1 \end{pmatrix} : a \in \mathbb Z\right\}$$
$$Ha = \left\{ \begin{pmatrix} 2 & a \\ 0 & 1 \end{pmatrix} : a \in \mathbb Z\right\}$$
Therefore $aH \subsetneq Ha$.
Najib Idrissi
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Let $G$ be the group of the maps $\Bbb R\to\Bbb R$ of the form $g(x)=ax+b,\ a\ne0$, and let $H$ be the subgroup with maps $h(x)=x+z,\ z\in\Bbb Z$.
Take a fixed $g\in G$, $g(x)=2x+b$. Then $gh(x)=2x+2z+b\in Hg$ since $Hg$ consists of the maps $x\mapsto 2x+b+n,\ n\in\Bbb Z$. But $hg(x)=2x+b+z$ and if this were in $gH$ for each $z\in\Bbb Z$, then there would be an $n\in \Bbb Z$ such that $hg(x)=2x+b+z=2x+2n+b$, thus $z=2n$, which is not possible for $z=1$. So $2x+b+1$ is in $Hg\setminus gH$.
Stefan Hamcke
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