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 $V$?
  • $\begingroup$ I think you are assuming that $\beta$ acts irreducibly in ${\rm GL}(2^{s-i},{\mathbb Z})$, but you have not said that. $\endgroup$
    – Derek Holt
    Mar 4, 2020 at 9:29
  • $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Mar 4, 2020 at 9:30
  • $\begingroup$ @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. $\endgroup$
    – usermath
    Mar 4, 2020 at 9:32
  • 1
    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Mar 4, 2020 at 9:35
  • 1
    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Mar 4, 2020 at 17:03

1 Answer 1


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 $Z(G)$.

  • $\begingroup$ 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. $\endgroup$
    – usermath
    Mar 5, 2020 at 9:44
  • $\begingroup$ 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. $\endgroup$
    – usermath
    Mar 5, 2020 at 9:51
  • 2
    $\begingroup$ Well if $H$ acts irreducibly then so does $G$ because $H < G$. $\endgroup$
    – Derek Holt
    Mar 5, 2020 at 9:54
  • $\begingroup$ I understand. Thanks. I have to check the paper. Possibly the case is different there. Any idea when the answer to 2 might be yes? Sorry, I am probably asking too many questions. $\endgroup$
    – usermath
    Mar 5, 2020 at 10:02

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