Show that the set of affine motions is a group. I was reading about group theory.  There was a problem given in the book:

An affine motion of $\mathbb {R}^2$ is a bijective mapping of the form $T_{Av}:\mathbb{R}^2\longrightarrow \mathbb{R}^2$ given by $T_{Av}(x)=A\underline{x} +\underline{v}$ for $x,v\in \mathbb{R}^2$ and $A\in GL(2,\mathbb{R})$ where $A\underline {x}$  is given by the left multiplication of the matrix $A$ with the column matrix $\underline {x}$. Show that the set of affine motions is a group under composition of mapping .

However, I am not getting what do they mean by $\underline{v}$? Is it a constant column matrix ? Also how to prove it ? I am not quite getting it...
 A: *

*Let $E$ be a real vector space. We define an affine application $t$ : it's an application $t:E\to E$ such that there exists $v\in E$, $a:E \to E$ linear, such that : $$\forall x \in E, t(x)=v+a(x)$$.

*Let's proof that $v$ and $a$ are uniquely determined by $t$. We necesserely have $v=t(0)$, where $0$ is the origin of $E$. Then, for $x\in E$, we necesseraly have $a(x)=t(x)-t(0)$. Q.E.D. We can now note $t_{a,v}$ but we won't because it is cumbersome.

3.We define an affine motion of $E$ as an affine bijective application.
4.We prove that $t$ is an affine motion iff $a$ is bijective.


*Let us come back to $E=\mathbb R ^2$. Let $v=(x_0,y_0)\in \mathbb R^2$. $\underline{v}=\begin{bmatrix}x_0 \\y_0\end{bmatrix}$.


*With the notations that I introduced, it should be easier to prove the claims.
A: With the notations from my previous answer,

*

*First, we will show that we have an internal composition law on the set $G$ of affine mappings from $\mathbb R^2$ to $\mathbb R^2$. Let $t_1=t_{a_1,v_1}$ and $t_2=t_{a_2,v_2}$ be two affine mappings. $\forall x\in \mathbb R^2, t_2(x)=v_2+a_2(x), t_1(t_2(x))=v_1+a_1(v_2+a_2(x))=v_1+a_1(v_2)+a_1(a_2(x))$ since $a_1$ is linear. So $t_1\circ t_2 \in G$

*$id_{\mathbb R^2}$ is the neutral element;

*$\circ$ is associative(associativity for mappings);

*Let $H$ be the set of affine motions; we know that $t=t_{a,v}$ is bijective iff $v\in GL(\mathbb R^2)$. Let $x, y\in \mathbb R^2. y=a+v(x)\iff y-a=v(x) \iff x=v^{-1}(y-a)\iff x=v^{-1}(y)-v^{-1}(a) \iff t^{-1}(y)=-v^{-1}(a)+v^{-1}(y) \iff t^{-1}=t_{-v^{-1}(a),v^{-1}}$. So $t$ has an inverse in $H$;

*From 1, 2, 3 and 4, we deduce that the set of affine motions is a group under composition of mappings .Q.E.D.

