Problem. Is there an example of a group $G$, a subgroup $H$ and an element $a \in G$ such that $|G : H| < \infty$ and $aH \subsetneq Ha$?
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1Do you have any ideas about this? For example, "clearly" $;G;$ must be infinite and so must $;H;$....why? Anything else, say about normality or stuff? – Timbuc Nov 21 '14 at 15:03
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Of course it is obvious. – Jihad Nov 21 '14 at 15:03
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Good you find that obvious. What else have you achieved so far? – Timbuc Nov 21 '14 at 15:04
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Group must be non-abelian xD. – Jihad Nov 21 '14 at 15:05
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Fine. Do you know any example of a non-abelian infinite group? If yes, do you know any subgroup in it which isn't normal? And if yes, can you show it has finite index? – Timbuc Nov 21 '14 at 15:06
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Good group is a group of matrices over, for instance $\mathbb{Z}$. – Jihad Nov 21 '14 at 15:10
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Well, we're now approaching something. You have a possible infinite non-abelian group (over $;\Bbb Z,,,,\Bbb Q;$ or whatever). What about finite index subgroups there? – Timbuc Nov 21 '14 at 15:12
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I found out that $\begin{pmatrix}a &b\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix}2 &0\ 0 & 1 \end{pmatrix} = \begin{pmatrix}2a &b\ 0 & 1 \end{pmatrix}$ but $\begin{pmatrix}2 &0\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix}a &b\ 0 & 1 \end{pmatrix} = \begin{pmatrix}2a &2b\ 0 & 1 \end{pmatrix}$ but unfortunately I can't find appropriate groups and subgroups. – Jihad Nov 21 '14 at 15:36
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I want to take $a = \begin{pmatrix}2 &0\ 0 & 1 \end{pmatrix}$ and $G = {\begin{pmatrix}a &b\ 0 & 1 \end{pmatrix}: a,b \in \mathbb{Z}}$ but it is not a group. – Jihad Nov 21 '14 at 15:41
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See also here and here. – Dietrich Burde Nov 21 '14 at 16:14
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Hint: $aH\subsetneq Ha\Rightarrow H\subsetneq a^{-1}Ha$.
Then, what can you say about $a^{-2}Ha^2$? Generalise.
If we relax the "finite index" assumption then there are lots of examples with $aH\subsetneq Ha$. Examples of such groups can be found in this question, and the links therein.
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...and apply finite index...(but be careful! this step is non-trivial. There is a possible issue. What is that issue? Why doesn't it matter?) – user1729 Nov 21 '14 at 15:41
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Oh, yes, for some $n$ $a^nH = H$ have to be equal to $Ha^n = H$ (since everything is finite). But it means that $H = a^{-n}Ha^n$. – Jihad Nov 21 '14 at 16:05
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Exactly! That is a better way of explaining it than I was thinking of - your explanation gets round my "possible issue". – user1729 Nov 21 '14 at 16:07