Find cosets and index of $GL_n$ and $SL_n$

Question

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.

Answer 1

Claim 1: Two matrices $g$ and $h$ are in the same right $SL_n$-coset if and only if $\det g=\det h$.

Claim 2: for $d\in \mathbb{R}^*$, $g={\rm diag}(d,1,\dots,1)$ is of determinant $d$.

Claim 3: $GL_n(\mathbb{R})/SL_n(\mathbb{R})=\mathbb{R}^*$.

Answer 2

Hint: Consider $\det :GL_n(\mathbb{R}) \to \mathbb{R}^*$. Then prove:

• $\det$ is a surjective group homomorphism

• $\ker \det = SL_n(\mathbb{R})$

• $GL_n(\mathbb{R})/SL_n(\mathbb{R}) \cong \mathbb{R}^*$