Cosets: Prove $\mathbf{H} \leq \mathbf{G}$ Let $\mathbf{G} = GL_n (\mathbb{R})$ and $\mathbf{H} = \{ H \in \mathbf{G}: \det(H) = \pm1 \}$.  
I'm trying to prove that
 $\mathbf{H} \leq \mathbf{G}$.
And then:
Given $A$, $B \in \mathbf{G}$, prove that $A\mathbf{H}=B\mathbf{H}$ if and only if $\det(A) = \pm \det(B)$.
Overall I've largely confused how the use of the determinant factors into the whole proof.
Any help is appreciated
 A: Presumably by $H \le G$ you mean that $H$ is a normal subgroup of
$G$ (more usually written $H \unlhd G$, I think)?
Let $\phi:G \to \mathbb{R}^+$ be given by $\phi(g) = | \det g |$. It is easy to check that $\phi(g_1 g_2) = \phi(g_1) \phi(g_2)$, so $\phi$ is a group
homomorphism.
Then $H = \ker \phi$, and so $H$ is a normal subgroup.
If $aH = bH$, then there are $h_1,h_2 \in H$ such that $a h_1 = b h_2$, and
since $\det(h_1), \det(h_2) \in \{-1,1\}$ we have $\det a = \pm \det b$.
If $\det a = \pm \det b$, then $\det(a^{-1} b)  \in \{-1,1\}$, and so $a^{-1} b \in H$. Noting that for any $h \in H$ we have $hH = H$, then $aH = a (a^{-1} b) H = b H$.
A: Well the easy direction : if $BH_1=AH_2$ For all $H_1,H_2$ in bold H.then obviously either  A and B have the same determinant of one is negative of the other one.
The other direction. Let $H$ be in that bold $H$ set. And assume $\det A=\pm\det B$
So $\det A^{-1}=\pm\frac{1}{\det B}$
Then $\det (A^{-1}BH)=\pm 1$ thus $A^{-1}BH$ Belongs to bold H for all H in bold H.
