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I am working on some basic context in Group Theory. I face to the following problem:

Let $G=GL(2,\Bbb Q)$ and $H\leq G$ be the subgroup of all diagonal invertible matrices. Find $$H_G=\cap_{g\in G}g^{-1}Hg$$

What I have done is to assume $H=\Big\{ \left( \begin{array}{ccc} a & 0 \\ 0 & b \\\end{array} \right)\mid ab \neq 0, a,b\in \Bbb Q\Big\}$ and take an arbitrary element $\left( \begin{array}{ccc} x & y \\ z & t \\ \end{array} \right) \in G$, and find the natural form of elements in $H_G$:

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But, clearly, I am on a wrong track. Please note me if you know applicable way for doing the core of $H$ in $G$. Thanks!

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Nice question...! –  amWhy Feb 18 '13 at 16:24

1 Answer 1

up vote 1 down vote accepted

We have for say $g = id$ that $gHg^{-1} = H$ so the question reduces to: which diagonal invertible matrices $\begin{pmatrix} c \\ & d \end{pmatrix}$ can be found in $gHg^{-1}$ for all $g$.

I claim that these are precisely $\lambda Id$, for $\lambda \in \mathbb{Q} - 0$.

To see this is necessary, if $x,y,z,t$ are all nonzero, for the resulting matrix $g^{-1} \begin{pmatrix} a \\ & b \end{pmatrix} g$ which you just computed to be diagonal, we need that $a = b =: \lambda$, which gives just those of the form $$\begin{pmatrix} \lambda \\ & \lambda \end{pmatrix}$$

To see this is sufficient, such a matrix $\lambda Id$ commutes with any other $g$, hence $g \lambda Id g^{-1} = \lambda Id$.

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Oh, thanks....I can see where I was confused in. Yes yes I see my defect. Thanks. –  Babak S. Jan 4 '13 at 19:51
    
So the core of $H$ will be $(\Bbb Q^*,.)$. Thanks –  Babak S. Jan 4 '13 at 19:52
    
No worries! It was a fun problem :) –  uncookedfalcon Jan 4 '13 at 19:53

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