Isomorphism of $\mathrm{GL}_3(\Bbb R)$ 
Prove that $\mathrm{GL}_3(\Bbb R)$ is isomorphic to $\Bbb R^{\times} \times \mathrm{SL}_3(\Bbb R)$. 

I read about and understand when using determinants that $\det\colon G\to R^{\times}$ is a homomorphism and that $\ker\det=S$ so $R^{\times}\cong G/\ker\det=G/S$. But I am struggling with this proof. Any idea where to start? My professor also gave us the hint that this is not true if you replace $3$ by $2$ and it would help if I knew what he was meant.
 A: If the isomorphism is correct, there is also a homomorphism from $GL_3(R)$ to $SL_3(R)$.  Think geometrically about what that is and eventually the rest becomes very simple.
A: Often, the easiest way to prove two things are isomorphic is to actually find an isomorphism between them.
Can you guess of a nice homomorphism from $\text{GL}_3(\mathbb{R})$ to $\mathbb{R}^\times \times \text{SL}_3(\mathbb{R})$? Or... maybe it's easier to find a nice homomorphism from $\mathbb{R}^\times \times \text{SL}_3(\mathbb{R})$ to $\text{GL}_3(\mathbb{R})$.
Once you have a good guess, see if you can find its inverse. If that exists and is a homomorphism too, you've proven them to be isomorphic.

I think you're trying to say that you already know that $\text{SL}_3$ is the kernel of the determinant homomorphism $\text{GL}_3 \to \mathbb{R}^\times$. There's a special trick that is often used in similar setups: to "split" the projection map $p:G \to G/H$. If you can find a homomorphism $s : G/H \to G$ with the property that $p \circ s$ is the identity, this can be used to construct the isomorphisms I mention above.
A key fact in this construction is that $x \cdot s(p(x))^{-1}$ is always an element of $H$.
