Prove $\operatorname{GL}_n(\Bbb{R})/\operatorname{SL}_n(\Bbb{R}) \cong \Bbb{R}^\times$ $$\operatorname{GL}_n(\Bbb{R})/\operatorname{SL}_n(\Bbb{R}) \cong \Bbb{R}^\times$$
This is trivial to prove with the first isomorphism theorem - by using the determinant as a endomorphism, then as $\operatorname{SL}_n(\Bbb{R})$ is $1$ under the determinant, it is the kernel, and by FIT, the above is proved.
I was wondering how to prove this without the isomorphism theorem ?
 A: To do it without FIT one should actually look at the cosets $g \cdot SL_n(\mathbb{R})$, where $g \in GL_n(\mathbb{R})$. I would take the set of representatives $$\{\textrm{sgn}(\alpha)|\alpha|^{1/n}I_n \cdot SL_n(\mathbb{R})| \hspace{1mm} \alpha \in \mathbb{R}^\times\}$$ Why is this a set of representatives? Well, all matrices in each coset have different determinants, so the cosets are certainly disjoint. They also make up all of $GL_n(\mathbb{R})$: if $g \in GL_n(\mathbb{R})$, then $\det g \neq 0$. Multiplying by $$\textrm{sgn}(\det g)|\det g|^{-1/n}I_n$$ gives some element in $SL_n(\mathbb{R})$ and so the claim follows.
Now, define the map $GL_n(\mathbb{R})/SL_n(\mathbb{R}) \to \mathbb{R}^\times$ by $\textrm{sgn}(\alpha)|\alpha|^{1/n}I_n \cdot SL_n(\mathbb{R}) \mapsto \alpha$, and one can easily check its an isomorphism.
I should comment that while this map is pretty neat, it depends on the choice of coset representatives, and is by no means canonical. I think using the first isomorphism theorem is a nicer way of doing things.
A: Hint: consider the homomorphism 
$$\mathbf{R}^\times \ni \lambda \mapsto (\lambda I_n) SL_n \in GL_n/SL_n$$
where $I_n$ is the $n\times n$ identity matrix and $gSL_n$ is the coset of $g$ in the quotient group.
