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Let $n$ be a positive integer and $G$ be a finite group of $n\times n$ matrices with integer coefficients, i.e. $G\subset\operatorname{GL}_n(\mathbb{Z})$. It is known that for sufficiently large $n$, the maximum order of such a group is $2^nn!$ by Feit (although relying on an unpublished manuscript of Weisfeiler I believe) and it is attained. I am interested in the coefficients that appear in the matrices of such a group. Specifically, I would like to know how many distinct numbers can appear.

Clearly it cannot be more than $2^nn!n^2$ if all coefficients of all matrices are distinct but I have the feeling that such a bound cannot be reached. By considering the companion matrix of a cyclotomic polynomial, I think it is possible to produce roughly $n$ or maybe even $n^2$ distinct coefficients. On the other hand, a group like the symmetric group can be encoded with $0,1$ matrices and the group that attains the $2^nn!$ upper bound only uses $-1,0,1$.

My questions are:

  • is it possible to derive a much better bound on the number of distinct coefficients, for example polynomial in $n$ ?

or

  • is there an example family of finite groups whose number of distinct coefficients is exponential in $n$ ?

I have the feeling that representation theory could help but I am not well-versed in the theory of finite groups.

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It is possible to get a number of distinct coefficients exponential in $n$. Here is an example.

Let $$B = \begin{bmatrix} 1 & -1 & -1 & -1 & \cdots & x_1 \\ 0 & 1 & 0 & 0 & & x_2 \\ 0 & 0 & 1 & 0 & \cdots & x_3 \\ 0 & 0 & 0 & 1 & & \vdots \\ & \vdots & & & \ddots & x_{n-1} \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$ We have $$B^{-1} = \begin{bmatrix} 1 & 1 & 1 & 1 & \cdots & -S \\ 0 & 1 & 0 & 0 & & -x_2 \\ 0 & 0 & 1 & 0 & \cdots & -x_3 \\ 0 & 0 & 0 & 1 & & \vdots \\ & \vdots & & & \ddots & -x_{n-1} \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$ where $S=x_1+x_2+\cdots+x_{n-1}$.

We consider the conjugate by $B$ of the group of diagonal $\pm 1$ matrices, of size $2^n$. Let $X$ be a diagonal matrix with $X_{i,i}=\varepsilon_i \in \{\pm 1\}$ and suppose $\varepsilon_n=1$. Then we can compute the upper right entry of $B^{-1}XB$ as $$-S + \sum_{i=1}^{n-1} \varepsilon_i x_i$$ If we take, for example, $x_i=2^i$, then these sums are all different for the $2^{n-1}$ possible choices of $X$, so at least $2^{n-1}$ different coefficients appear in the matrices of the group.

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