# If $A$ is a free abelian normal subgroup, and $K'$ has a unipotent image then $C_A(K')$ is non-trivial

Here's the context of the question:

Let $$G$$ be an infinite polycylic group. It is known that there is $$A \triangleleft G$$ with $$A \cong \mathbb{Z}^d$$ for $$d >0$$.

Define a homomorphism $$\phi: G \to$$ Aut$$(A) \cong GL(d, \mathbb{Z})$$ by $$\phi(g)(a) = g^{-1}ag$$.

Since $$G$$ is solvable $$\phi(G)$$ is as well, and as it is a subset of a linear group, there is $$L \leq \phi(G)$$ triangularizable, finite index and normal.

Then $$[L,L] := L'$$ is unipotent and so $$(L' - 1)^d = 0$$.

Denote $$K = \phi^{-1}(L)$$ so that $$K' = \phi^{-1}(L')$$.

Then let the centralizer $$C_A(K') = \{a \in A :\forall k'$$ in $$K'$$ $$k'a = ak'\}$$.

Question: does it follow that $$C_A(K')$$ is non trivial in this case?

I can't see why.

• If $e$ is minimal with $(L'-1)^e=0$, then ${\rm im}((L'-1)^{e-1}) \le C_A(K')$. – Derek Holt Nov 25 '18 at 20:01