Let $GL_n(\mathbb{R})$ and $SL_n(\mathbb{R})$ be the general linear and special linear groups.
I want to find the right cosets of $H=SL_n$ in $G=GL_n$ and the index $[GL_n:SL_n]$.
The right cosets will be given by $Hg=\{M_HM_G: M_H\text{ is a matrix in }SL_n\}$.
Would it be right to say that the cosets are given by the set $GL_n$; the argument I have in mind hs the fact that if $M_H\in SL_n$ then $\det(M_H)=1$ and since $\det(AB)=\det(A)\det(B)$ follows $\det(M_HM_G)=\det(M_G)$. The $GL_n$ set has $n\times n$ invertible matrices, and matrices are invertible if the determinant is non zero, which applies for every $M_HM_G$ because $\det(M_G)$ is not zero.
And isn't the index infinite? Because of the cardinal of $\mathbb{R}$ and this theorem Number of left cosets of the special linear group in the general linear group.
Best Answer
Hint: Consider $\det :GL_n(\mathbb{R}) \to \mathbb{R}^*$. Then prove:
$\det$ is a surjective group homomorphism
$\ker \det = SL_n(\mathbb{R})$
$GL_n(\mathbb{R})/SL_n(\mathbb{R}) \cong \mathbb{R}^*$