Question: Let $N=\operatorname{GL}\_n(\mathbb{R})$ and $M=\operatorname{SL}\_n(\mathbb{R})$. Is $M$ normal subgroup in $N$? Why or why not?

I know how to do this with $\operatorname{GL}\_2(\mathbb{R})$ and $\operatorname{SL}\_2(\mathbb{R})$ but with $N=\operatorname{GL}\_n(\mathbb{R})$ and $M=\operatorname{SL}\_n(\mathbb{R})$ I don't even know all of their elements so I can't check the left and right cosets of $M$ in $N$.

Answer: 
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....?