I was reading an algebra paper, and the problem that appeared to me is the following:
The authors defined a group $G = A \rtimes \left<x\right>$, where $A$ is a finitely generated free abelian group and $\left<x\right>$ is the infinite cyclic group and such that $x$ does not normalize any non-trivial cyclic subgroup of $A$. Then the element $x$ acts on $A$ in the following way for suitable integers $t_{i,j}$, given that $\{a_{1}, ..., a_{r}\}$ is the independent set that generates $A$:
$$x^{-1}a_{i}x = \prod\limits_{j}a_{j}^{t_{i,j}}.$$
Now, the authors defined the matrix $T = [t_{i,j}] \in GL_{r}(\mathbb{Z})$.
The characteristic polynomial $f(y) = \det(yI - T)$ is such that $f(0) = \det(T) = \pm 1$ (because $[t_{i,j}] \in GL_{r}(\mathbb{Z})$) and $f(1) \neq 0 \neq f(-1)$ - this part is ok, given that $x$ does not normalize non-trivial cyclic subgroups of $A$. Now the authors also say that $f(y)$ is not the power of a linear polynomial, "since the composition factors of $x$ on $\mathbb{Q} \otimes A$ cannot all be linear".
I simply could not understand this last part, and I don't know how to justify the fact that $f(y)$ is not the power of a linear polynomial. What does this last sentence mean after all? If anyone could provide some sort of insight I would be really grateful.
Thanks in advance.
Edit: By definition, an element $x \in G$ normalizes a subgroup $H \leq G$ iff $xHx^{1} = H$.
Suppose that $f(y)=(X-a)^n$. It implies that The map $L_x:A\rightarrow A$ defined by $L_x(u)=x^{-1}ux$ verifies $(L_x-aId_A)^n=0$. This implies that $Ker(L_x-aId_A)\neq 0$, we deduce that there exists $u\in A, u\neq 0$ with $xux^{-1}=au$. This implies that $L_x$ normalize the non trivial subgroup generated by $x$. Contradiction.