I asked this question in Math StackExchange. I will be really grateful for any help here.

I was solving a problem and in the middle, I came across this. In the following, we fix integers $s\ge 2$ and $0\le i\le s-1$. We set-up the following.

- $\beta\in \rm{GL}(2^{s-i},\mathbb Z)$ is such that $\beta\ne I_{2^{s-i}}$ and order of $\beta$ is $2^{s-i+1}$.
- $ c=\begin{pmatrix} \beta & 0 \\ 0 & I_{2^s-2^{s-i}} \\ \end{pmatrix} \in \rm{GL}(2^{s},\mathbb Z). $
- $P$ is the $i$-fold wreath product of the permutation group $\langle (1,2) \rangle$.
- Let $U\le W=\langle c \rangle \wr P\le \rm{GL}(2^s,\mathbb Z)$
- $Y=I_{2^i}\otimes\langle\beta\rangle=Z(W)$ is a cyclic subgroup of $W$.

Suppose $Z(U) < Y=Z(W)$. We consider the decomposition $V=\mathbb Q^{2^s} = V_1 \oplus \ldots \oplus V_{2^i}$ according to $Y$ that is this is a decomposition into the direct sum of irreducible $Y$-modules. Since $U \le W$, the group $U$ permutes the direct summands $V_j$ and the kernel of this action thus contains $Z(U)$.

My question is:

- Since $Z(U)$ is cyclic (as it is a subgroup of the cyclic group $Y$), is it true that then $Z(U)$ acts reducibly on each $V_j$?
- If 1. is true then does $U$ act reducibly on $V$?