Is $SL_n(\mathbb{R})$ a normal subgroup in $GL_n(\mathbb{R})$?
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 $N$.
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 $\det(AB) = \det(A)\det(B)$.
Hint:
Is the determinant map
$$\det:GL_n(\Bbb R)\to\Bbb R^*\,$$
a homomorphism....?