Commutator subgroup of a group of affine transformations This problem is taken from Berkeley problems in Mathematics. The solution is not given in the book.
Problem: Let $T$ be an invertible linear transformation $V \to V$.
Let $G$ be the group of maps $f_{k, a}$ defined by $f_{k, a} = T^{k}(x)+a$ where $k, a$ vary over the integers and $V$ respectively.
Prove that the commutator subgroup $G'$ of $G$ is isomorphic to the additive group $(T-I)V$, the image of $T-I$.
So, far I have only been able to compute a generic element $ghg^{-1}h^{-1}$ where $g = f_{k_2, a_2}$ and $h = f_{k_1, a_1}$ and it looks like $x - (a_1 - a_2) - T^{k_1}(a_2) + T^{k_1 + k_2}(a_1)$.
From here I dont know how to show the isomorphism. It is not clear to me what an element like above would correspond to in $(T-I)V$?
I am looking for hints.
Update: Above computation of $ghg^{-1}h^{-1}$ is wrong as pointed out conceptually by Arturo.
The correct formula as can be easily checked is $x+(a_2 -a_1) -T^{k_1}(a_2) + T^{k_2}(a_1)$.
 A: This is Problem 7.4.11.
As I noted in the comment, your computation of $ghg^{-1}h^{-1}$ is necessarily incorrect. You can spot this beause $[h,g]=[g,h]^{-1}$, so if we reverse the roles of $g$ and $h$, we should get the inverse transformation to the one you've calculated. However, exchanging the roles of $g$ and $h$ in your calculations would yield
$$x - (a_2-a_1)-T^{k_2}(a_1)+T^{k_2+k_1}(a_2),$$
which is not the inverse of the transformation you've calculated. So there is definitely something wrong in your calculation.
By my calculation, we have:
$$\begin{align*}
f_{k,a}\circ f_{\ell,b}(x) &= f_{k,a}(T^{\ell}(x)+b)\\
&= T^{k+\ell}(x)+T^{k}(b)+a\\
&= f_{k+\ell,T^{k}(b)+a}\\
(f_{k,a})^{-1} &= f_{-k,-T^{-k}(a)}\\
(f_{k_2,a_2}\circ f_{k_1,a_1})\circ(f_{k_2,a_2})^{-1}\circ(f_{k_1,a_1})^{-1} &= f_{k_2+k_1,T^{k_2}(a_1)+a_2}\circ f_{-k_2,-T^{-k_2}(a_2)}\circ f_{-k_1,-T^{-k_1}(a_1)}\\
&= f_{k_2+k_1,T^{k_2}(a_1)+a_2}\circ f_{-k_2-k_1,-T^{-k_2-k_1}(a_1)-T^{-k_2}(a_2)}\\
&= f_{0,-a_1-T^{k_1}(a_2)+T^{k_2}(a_1)+a_2}.
\end{align*}$$
So the commutator results in the function
$$x\longmapsto x + (a_2-a_1)-T^{k_1}(a_2)+T^{k_2}(a_1).\tag{1}$$
Note that if we exchange roles of $g$ and $h$, we will obtain the function
$$x\longmapsto x + (a_1-a_2) - T^{k_2}(a_1)+T^{k_1}(a_2).\tag{2}$$
and indeed, the function $(2)$ is the inverse of the function $(1)$.
Now, consider the commutator you get when $g=f_{k_1,a_1}$, and $h=f_{1,0}$. The commutator we get is $f_{0,a_1-T(a_1)}$.
I claim that every commutator can be obtained as a composition of functions of the form $f_{0,v-T(v)}$. Indeed, as we just saw, every such function is a commutator. Now note that the commutator as we calculated above is equal to
$$f_{0,a_2-T^{k_1}(a_2)}\circ f_{0,-a_1-T^{k_2}(-a_1)}.$$
So it suffices to show that every function of the form $f_{0,v-T^k(v)}$ is a composition of functions of the form $f_{0,v-T(v)}$.
We proceed by induction on $k$. The result is immediate if $k=1$. Assuming we can obtain the function $f_{0,v-T^k(v)}$, we have:
$$f_{0,v-T^k(v)}\circ f_{0,T^{k}(v)-T(T^k(v))} = f_{0,v-T^k(v)+T^{k}(v)-T^{k+1}(v)} = f_{0,v-T^{k+1}(v)}.$$
So we are done.
This proves that the commutator subgroup $G'$ is generated by the functions $f_{0,v-T(v)}$. Taking inverses, we get the functions $f_{0,T(v)-v} = f_{0,(T-I)(v)}.$
Now note that
$$f_{0,(T-I)(v)} \circ f_{0,(T-I)(w)} = f_{0,(T-I)(v+w)}.$$
This suggests the desired isomorphism.
