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I've been struggling to show that $\mathrm{SL}_2(\mathbb{R})$ is a normal subgroup of $\mathrm{GL}_2(\mathbb{R})$. I already proved that $\mathrm{SL}_2(\mathbb{R})\leq\mathrm{GL}_2(\mathbb{R})$ (not shown). Now I want to show that $$ A\cdot \mathrm{SL}_2(\mathbb{R})=\mathrm{SL}_2(\mathbb{R})\cdot A $$ for every $A\in \mathrm{GL}_2(\mathbb{R})$.

I know that $\det(AB)=\det(A)\det(B)=\det(B)\det(A)=\det(BA)$. Thus, $$\det(A\cdot \mathrm{SL}_2(\mathbb{R}))=\det(\mathrm{SL}_2(\mathbb{R})\cdot A )$$ but this does not seem to help me prove normality.

I thought that perhaps rearranging in the following form would help:

$$ A\cdot \mathrm{SL}_2(\mathbb{R})\cdot A^{-1}=\mathrm{SL}_2(\mathbb{R}) $$ If I can show that $A\cdot \mathrm{SL}_2(\mathbb{R})\cdot A^{-1}$ has determinant 1, then I am done. How can I do this?

I would like a hint (no full solutions, please) on how I can proceed.

Thanks!

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    $\begingroup$ Kernels of group homomorphisms are always normal. (Can you prove it?) Also it doesn't make sense to say a set has determinant $1$; you mean all of its elements have determinant $1$. You're trying to show $ABA^{-1}$ has determinant $1$ given $\det B=1$. Have you tried simply taking the determinant and seeing what happens? You said you already know that $\det(XY)=\det(X)\det(Y)$; use it. $\endgroup$
    – whacka
    Feb 4, 2015 at 16:40
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    $\begingroup$ Hang on... I think I got it. $\det(ABA^{-1})=\det(A)\det(B)\det(A^{-1})=\det(A)\det(A^{-1})\det(B)= \det(I)\det(B)=\det(B)=1$ (because $B\in \mathrm{SL}_2(\mathbb{R})$). Thus, $A\cdot \mathrm{SL}_2(\mathbb{R})\cdot A^{-1}\in\mathrm{SL}_2(\mathbb{R})$. How can I show that equality comes from this? (instead of just $\in$) $\endgroup$ Feb 4, 2015 at 16:45
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    $\begingroup$ Something about your notation is telling me that you are trying to treat $\mathrm{SL}_2(\Bbb{R})$ as a matrix, when it's actually a set of matrices. What you are trying to prove is that $ABA^{-1}\in\mathrm{SL}_2(\Bbb{R})$ when $B$ has det 1. $\endgroup$ Feb 4, 2015 at 16:46
  • $\begingroup$ Sure. You could have used $\det B=1$ earlier in the calculation, but yes. Well, except you're using the $\in$ symbol when you mean $\subseteq$. $\endgroup$
    – whacka
    Feb 4, 2015 at 16:47
  • $\begingroup$ I have been referring to $\mathrm{SL}_2(\mathbb{R})$ as the set matrices. Edit: I now see that it is $\subseteq$--obviously. They are all sets. $\endgroup$ Feb 4, 2015 at 16:48

3 Answers 3

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Hint: Can you write $SL_2$ as a kernel? You certainly know some multiplicative maps from linear algebra. (From my experience, the easiest way to show that some subgroup is normal is to exhibit it as a kernel of a homomorphism.)

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  • $\begingroup$ It seems that $\mathrm{SL}_2(\Bbb{R})=\mathrm{Ker}(f)$ where $f(A)=\det(A)$ $\endgroup$ Feb 7, 2015 at 13:24
  • $\begingroup$ Indeed, this is the definition of $SL_2$! $\endgroup$ Feb 7, 2015 at 13:25
  • $\begingroup$ How does the kernel relate to normality? Would I use the first isomorphism theorem (I need to go look it up in my book--I don't have it memorized)? $\endgroup$ Feb 7, 2015 at 13:26
  • $\begingroup$ @PatrickShambayati. If $f:G\rightarrow H$ is a homomorphism of groups, then its kernel is a normal subgroup of $G$. Try to prove this, it just takes one line, you merely have to check the definition of a normal subgroup. $\endgroup$ Feb 7, 2015 at 13:28
  • $\begingroup$ If $a\in\mathrm{ker}(f)$ then $f(xax^{-1})=f(x)f(a)f(x^{-1})=f(x)f(x^{-1})=f(x)f(x)^{-1}=e_H$ for all $x\in G$. Thus, $xax^{-1}\in\mathrm{ker}(f)$ and $\mathrm{ker}{f}\trianglelefteq G$ $\endgroup$ Feb 26, 2015 at 17:40
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Hint: If $A \in {\rm SL}(n, \Bbb R)$ and $G \in {\rm GL}(n, \Bbb R)$, you want to prove that $G^{-1}AG \in {\rm SL}(n, \Bbb R)$. But: $$\det(G^{-1}AG) = \det(G^{-1})\det A\, \det G.$$

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Hint: determinants are multiplicative, and real numbers commute.

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