Showing a translation group is a normal subgroup of an affine group Let V be an n-dim vector space over the field $\mathbb{F}.$
$A \in GL\left ( n,\mathbb{F} \right )$ and $v \in V$
Define the affine transformation $t_{A,v}$:
$V\rightarrow V$
$x \mapsto xA+v$

Affine general linear group
$AGL\left ( n,\mathbb{F} \right )=\left \{ t_{A,v} \mid A \in GL\left ( n,\mathbb{F} \right ),v \in V\right \}$

Verify that
$\left ( T,+ \right )=\left \{ t_{I,v} \mid v \in V\right \}$ is the normal subgroup of all translation.
Check first that $\left ( T,+ \right )$ is a subgroup.
It is easy to show that the identity matrix I exists.
Also, T is closed under permutation.
since $\left ( \left ( x \right )^{t_{I,v}} \right )^{I,w}=\left ( xI+v \right )^{t_{I,w}}=\left ( xI+v \right )I+w=xII+vI+w=x+\left ( v+w \right )=\left ( x \right )^{t_{I},v+w}$
Not quite sure about inverse translation but the inverse translation is
a map from the element xA+v to x.
Also, I am interested in verifying that $\left ( T,+ \right )$ is a normal subgroup of $AGL\left ( n,\mathbb{F} \right ).$

Recall that a subgroup H of group G is a normal subgroup $\mathbf{IFF} Hg=gH \forall g \in G$.

We require that$ \left ( \left ( x \right )^{t_{I,v}} \right )^{t_{A,v}}=\left ( \left ( x \right )^{t_{A,v}} \right )^{t_{I,v}}$
So, $\left ( \left ( x \right )^{t_{I,v}} \right )^{t_{A,v}}=xA+vA+v$
At this point, it doesn't as though translation can be fulfilled.
Any help is appreciated. Thanks in advance.
 A: You misinterpreted the requirement $Hg=gH$ a little bit. What it requires is that the two sets 
$$
Hg=\{hg\mid h\in H\}
$$
and
$$
gH=\{gh\mid h\in H\}
$$
should be the same set for any fixed element $g\in G$. But that does not mean that $gh$ should be equal to $hg$ for all $h\in H, g\in G$. What it does mean is that $gh$ is an element of the set $Hg$ for all $h\in H, g\in G$. In other words, it means that there has to be an element $h'\in H$ (possibly depending on both $h$ and $g$) such that $gh=h'g$.
The equation
$$
gh=h'g\qquad(*)
$$
actually determines $h'$ uniquely. More specifically multiplying $(*)$ from the right be $g^{-1}$ implies that $h'=ghg^{-1}$. This leads us to the other common criterian for normality: a subgroup $H\le G$ is normal iff $ghg^{-1}\in H$ for all $g\in G, h\in H$.
So in your case you need to check that composition
$$
t_{A,w}\circ t_{I,v}\circ t_{A,w}^{-1}
$$
is equal to some translation $t_{I,u}$ for some vector $u$ that does not have to be equal to $v$. Alternatively, you need to produce a vector $u$ (that may depend on $v,w$ and $A$, but must not depend on $x$) such that
$$
(x^{t_{I,v}})^{t_{A,w}}=(x^{t_{A,w}})^{t_{I,u}}.
$$

Another mistake I spotted is that you used the same vector $v$ in both elements $t_{A,v}$ and $t_{I,v}$. This is not general enough, because you are then not checking condition $(*)$ for all $g\in G$. That's why I used $v$ and $w$ in the previous paragraph.
