Affine group, semi-direct product and linear transformations According to wikipedia the Affine group is the semi-direct product of a vector space $V$ and the general linear group $GL(V)$. Here is the definition of the semi-direct product in terms of matrices and vectors as given in the wikipedia link:
\begin{equation}
  (M_1, v_1)\cdot(M_2, v_2) = (M_1 M_2, v_1 + M_1v_2)
\end{equation}
By definition, I know the affine group is the set of all invertible affine maps and I know that the general linear group is the set of all invertible linear maps. I also know that an affine map is just a linear map without preserving the origin, and hence all linear maps are affine maps. 
It seems reasonable to me that affine maps can be constructed from linear maps, but I do not see everything from the above definition of semi-direct product. The only other idea I have is relating the difference between linear maps and affine maps, which is: translations. So I am thinking that the second coordinate vector gives us our translation, and the first coordinate gives us our linear transformation, however why not switch $v_1$ and $v_2$ in the second coordinate? Or why is $M_1$ in the second coordinate and not $M_2$?
Specifically,


*

*How does the above semi-direct product give us all invertible affine maps?

*Why not just a direct product or some other product? 

*Why is the equation exactly as specified? Why not switch $v_1$ and $v_2$ in the second coordinate or $M_1$ and $M_2$? I guess the second coordinate confuses me the most since it is what makes it different then a direct product, so please explain why the second coordinate takes that form.

*Please share anything else you think is useful in understanding semi-direct products, how they arise (constructing groups?) and how they work in this specific example.

*Please let me know where I am right and wrong :)


Thanks for all the help 
 A: Here are a few quick commnents. Let $T$ be the group of translations and $G = {\rm GL}(V)$, and note that their intersection is trivial. You can show geometrically that every affine transformation can be decomposed (uniquely) as a composite of a linear map fixing the origin followed by a translation.
So all elements have a decomposition as $tg$ with $t \in G$ and $g \in G$, and hence the affine group $A$ is a product $TG$ with $T \cap G = \{1\}$. Now you can check that $T$ is a normal subgroup of $A$ but $G$ is not. This is exactly when we get a semidirect product. For a direct product you would need both $T$ and $G$ to be normal in $A$.
The multiplication rule they give is just the way it happens to be. But you could equally well write the elements as $(v,M)$ rather than as $(M,v)$, which I find more natural, because it fits better with the decomposition $gt$. The multiplication would then work out as $(v_1,M_1)(v_2,M_2) = (v_1+M_1v_2,M_1M_2)$, which I find more transparent, because it is the $M_1$ moving rightwards past the $v_2$ that is causing the $v_2$ to get replaced by $M_1v_2$ in the product. This is the way multiplication works in semidirect prodcuts in general.
