Let \(T:V\to V\) be an invertible linear transformation of a vector space \(V\). Denote by \(G\) the group of all maps \(f_{k,a}:V\to V\) where \(k\in\ZZ\), \(a\in V\), and for \(x\in V\)
\[ f_{k,a}(x)=T^kx+a\quad(x\in V). \]
Prove that the commutator subgroup \(G'\) of \(G\) is isomorphic to the additive group of the vector space \((T-I)V\), the image of \(T-I\). (\(G'\) is generated by all \(ghg^{-1}h^{-1}\), \(g\) and \(h\) in \(G\).)
Solution
Write \(f_{k,a}\) as the pair \((k,a)\). Composition gives \[ (k,a)(\ell,b)=(k+\ell,\ a+T^k b), \] and therefore \[ (k,a)^{-1}=(-k,-T^{-k}a). \] Let \[ W=(T-I)V \qquad\text{and}\qquad N=\{(0,w):w\in W\}. \] Since \((0,w)(0,w')=(0,w+w')\) and \(W\) is a subspace, \(N\) is a subgroup of \(G\).
1. \(G'\subseteq N\).
For \(g=(k,a)\) and \(h=(\ell,b)\), direct multiplication gives \[ \begin{aligned} [g,h] &=(k,a)(\ell,b)(k,a)^{-1}(\ell,b)^{-1}\\ &=\bigl(0,(I-T^\ell)a+(T^k-I)b\bigr). \end{aligned} \] For every integer \(n\), one has \[ (T^n-I)V\subseteq (T-I)V. \] Indeed, for \(n>0\), \[ T^n-I=(T-I)(I+T+\cdots+T^{n-1}), \] while for \(n<0\), \[ T^n-I=-T^n(T^{-n}-I), \] and the positive-power factorization applies to \(T^{-n}-I\); all powers of \(T\) commute with \(T-I\). Thus both \((I-T^\ell)a\) and \((T^k-I)b\) lie in \(W\), so every commutator lies in \(N\). Since \(N\) is a subgroup, \(G'\subseteq N\).
2. \(N\subseteq G'\).
Let \(w\in W\). Then \(w=(T-I)v\) for some \(v\in V\). Take \[ g=(1,0),\qquad h=(0,v). \] The commutator formula in the proof of step 1 gives \[ [g,h]=(0,(T-I)v)=(0,w). \] Hence every element of \(N\) is a commutator and lies in \(G'\).
3. The map\[ \Phi:(T-I)V\longrightarrow G',\qquad \Phi(w)=f_{0,w}, \]is an isomorphism from the additive group of \((T-I)V\) onto \(G'\), so\[ \boxed{G'\cong ((T-I)V,+)}. \]
Steps 1 and 2 give \[ G'=N=\{f_{0,w}:w\in (T-I)V\}, \] so \(\Phi\) takes values in \(G'\) and is surjective. Since \[ f_{0,w}\circ f_{0,w'}=f_{0,w+w'}, \] \(\Phi\) is a homomorphism. It is injective because \(f_{0,w}(0)=w\).
4. Q.E.D.
Step 3 is the required isomorphism.