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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.

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

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