# Cyclic group acting reducibly on $\mathbb Q^{2^s}$

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:

1. 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 $$V_j$$?
2. If 1. is true then does $$U$$ act reducibly on $$V$$?
• I think you are assuming that $\beta$ acts irreducibly in ${\rm GL}(2^{s-i},{\mathbb Z})$, but you have not said that. – Derek Holt Mar 4 at 9:29
• I think you could have $\beta$ (and hence $Y$) of order $2m$ with $m$ odd and $Z(U)$ of order $m$ with $Z(U)$ still acting irreducibly. – Derek Holt Mar 4 at 9:30
• @DerekHolt I think in my case order of $\beta$ is a power of $2$, does that help in any way? Also $\beta$ is not identity matrix. I will edit the question. But assuming this can we say something? Any help will be appreciated. – usermath Mar 4 at 9:32
• If $\beta$ has order $2^k$ and acts irreducible, then $s-i = k-1$ and the answer to Question 1 is yes. I haven't thought about Question 2. – Derek Holt Mar 4 at 9:35
• For each $m> 0$, up to equivalence there is a single faithful irreducible rational representation of the cyclic group of order $m$, and this has dimension $\Phi(m)$. You can take the image of a group generator to be the companion matrix of the $m$-th cyclotomic polynomial. For $m=2^k$, this polynomial is $x^{2^{k-1}}+1$. So any proper subgroup acts reducibly in this case. This effectively proves 1. – Derek Holt Mar 4 at 17:03

The answer to Question 2 is no, and I found a counterexample using computer calculations. We take $$s=2$$, $$i=1$$, and $$\beta = \left(\begin{array}{rr}0&1\\ -1&0\end{array}\right)$$, so $$\beta$$ has order $$4$$, and $$G = \langle x,y \rangle$$ with $$x = \left(\begin{array}{rrrr}0&1&0&0\\-1&0&0&0\\0&0&1&0\\0&0&0&1\end{array}\right),\ y = \left(\begin{array}{rrrr}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{array}\right).$$ Let $$H = \left\langle\,\left(\begin{array}{rrrr}0&-1&0&0\\1&0&0&0\\0&0&0&1\\0&0&-1&0\end{array}\right), \ \left(\begin{array}{rrrr}0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0\end{array}\right)\,\right\rangle = \langle y^xy,x^2y \rangle.$$ Then $$H$$ has order $$8$$ (isomorphic to $$Q_8$$), it acts irreducibly, and its centre has order 2 and is a proper subgroup of $$Z(G)$$.
• Thanks a lot. Just wondering, $H$ acts irreducibly but don't we need to decide whether $G$ acts reducibly or not? Really sorry if I missed something. – usermath Mar 5 at 9:44
• If you are thinking about the origin of the question, I was reading the paper "Determination of the uniserial space groups with a given coclass" by Bettina Eick. The claim of reducibility of $U$ is made in p. 630 (in the proof of Theorem 19a). I have a similar situation, so wondering how the claim is made there. – usermath Mar 5 at 9:51
• Well if $H$ acts irreducibly then so does $G$ because $H < G$. – Derek Holt Mar 5 at 9:54