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

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.

• We can use dimensional counting. How many parameters do we need to specify a matrix in $GL_n$? $n^2$, and for $SL_n$? we have the same number of parameters and one restriction, hence we get $n^2-1$ parameters. Therefore, the factor space is one dimensional. More precisely, it can be realized as $\mathbb{R}\setminus 0=\{\text{directed volume of } AB_1, \text{ where }A\in GL_n \text{ and } B_1 \text{ is the unit ball} \}$. I hope it helps. – user90189 May 10 '18 at 0:46
• @user90189 I'm sorry but I don-t quite understand your answer. Particularly after "it can be realized as" – Cure May 10 '18 at 3:07

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

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

• I could work out how $g=diag(d,1,...1)$ is the form of the elements that we will use to find the cosets of $SL_n$ in $GL_n$ and hence use claim 1 and 2. But could you please elaborae on claim 3? – ali Aug 12 '20 at 17:38
• @ali, yes these elements are representative of the cosets. Claim 3 may be proved using Ihf's answer (the surjectivity is proved using my Claim 2). – Clément Guérin Aug 13 '20 at 17:14

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

• $\det$ is a surjective group homomorphism

• $\ker \det = SL_n(\mathbb{R})$

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