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Let $N= GL_n(\mathbb{R})$ and $M=SL_n(\mathbb{R})$. Is $M$ normal subgroup in $N$? Why or why not?

I know how to do this with $GL_2(\mathbb{R})$ and $SL_2(\mathbb{R})$ but with $N= GL_n(\mathbb{R})$ and $M=SL_n(\mathbb{R})$ I don't even know all of their elements so I can't check the left and right cosets of $M$ in $N$.

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$det (ABA^{-1})= det B$ – i707107 Apr 13 '13 at 0:25
up vote 6 down vote accepted

Remember that $H$ is normal in $G$ if $ghg^{-1} \in H$ for all $g \in G$, $h \in H$. Also note that for square matrices $A, B$ we have $\det(AB) = \det(A)\det(B)$.

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Is the determinant map

$$\det:GL_n(\Bbb R)\to\Bbb R^*\,$$

a homomorphism....?

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I haven't learn about homomorphism yet – Diane Vanderwaif Apr 13 '13 at 1:39
Then I can't understand how you're dealing with this kind of questions...but perhaps it's only me and my way of seeing the order of teachind and learning stuff. – DonAntonio Apr 13 '13 at 1:40
@DonAntonio Since for any element in the special group, $\det A=\det A^{-1}=1$, $\det AB=\det A\det B$ means conjugations by elements of the special group preserve determinants. It would have been more useful to write that instead of resorting to homomorphisms, in my opinion. – Pedro Tamaroff Apr 13 '13 at 1:51
I'm with @DonAntonio on this one. Maybe a "homomorphism" is a more abstract concept than a normal subgroup, but it's difficult understanding why you would care about normal subgroups without knowing about homomorphisms. – user29743 Apr 13 '13 at 4:41
(at the OP - this isn't meant as anything negative about you, of course! Just a comment that it's frustrating when things are presented in this order.) – user29743 Apr 13 '13 at 4:42

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