Describe the finite order integer matrices over complex field

Question

I stuck so much on this question!

I need to describe finite order integer matrices without 1 eigenvalues over $\mathbb C$. I need description in terms of classes of equivalent matrices(such that exist complex matrix C such that $C^{-1}AC = B$)

I came to this results. Let A be the integer matrix of order $m$. First of all A can be diagonalized so is $diag(\lambda_1, ..., \lambda_n)$ in some basis. If $\lambda$ is its eigenvalue, then $\lambda^{m} = 1$.

The minimal polynomial $p$ for A divides $x^m - 1$ and belongs to $\mathbb Z[x]$ because $A \in Mat(\mathbb Z)$. Actually $p(\lambda_i) = 0$. So this should give some limitations on $(\lambda_1, ..., \lambda_n)$. I need to get all possible cases for $(\lambda_1, ..., \lambda_n)$. My hypothesis is "$(\lambda_1, ..., \lambda_n)$ splits into groups where every group is all roots of $\frac{x^k - 1}{x-1}$".

Edit 1. I was really inaccurate with formulating my hypothesis. I really mean that $(\lambda_1, ..., \lambda_n)$ splits into groups where every group is all roots of some irreducible polynomial in decomposition of $\frac{x^k - 1}{x-1}$

Edit 2. I'm sorry if I messed you up, I'll try to give much more readable description of my question. I'm doing a research and found out that for some group all automorpmism are in 1-1 correspondence with the classes of simmilar integer matrices of finite order without 1-eigenvalues. So what I want is to somehow enumerate this classes. I divided the problem in two parts:

1) Prove that the set of classes(that I described above) is finite

2) Give a constructive way to enumerate this classes

Hope now my question is more readable. Any help is appreciated!

Answer 1

$A,B$ are not equivalent matrices but similar matrices. It's hard to understand what you're looking for (is it a homework ?); the integer matrices with finite order are well understood (thanks to Minkowski, Taussky and Todd). cf. for example

https://pdfs.semanticscholar.org/d72f/50b01413336e0c2b4f01859b5c39e01d7ad1.pdf

Of course, here we can have $1\in spectrum(A)$; you have to work a little; I hope this will not put your life at stake.

Let $\alpha_n=\{m;$ there is $A\in M_n(\mathbb{Z}) \;s.t.\;A^m=I_n\}$ and $s_n=\max(\alpha_n)$. Some results:

1. Let $m=p_1^{e_1}\cdots p_t^{e_t}$ be the decomposition of the integer $m$ on the prime factors $p_1<\cdots<p_t$. Then $m\in\alpha_n$ iff

when $p_1^{e_1}=2$: $\sum_i(p_i-1)p_i^{e_i-1}-1\leq n$

otherwise: $\sum_i(p_i-1)p_i^{e_i-1}\leq n$

2. $\alpha_{2k+1}=\alpha_{2k}$.

3. $\alpha_2=\alpha_3=\{2,3,4,6\}$, $\alpha_4=\alpha_5=\{2,3,4,5,6,8,10,12\}$.

4. $s_{22}=2520$. Note that $2520\in\alpha_{22}$ because $2520=2^3.5.7.9$ and $2^2+4+6+8=22$.