TITLE: Construct a rational matrix $A$ s.t. $A^m = I$
QUESTION [5 upvotes]: Let $K$ be a field of either $\mathbb{C}$, $\mathbb{R}$ or $\mathbb{Q}$, Let $V$ be a $n$ dimensional vector space over $K$.
I want to construct a matrix $A \in GL(V)$ s.t. $A^m = I$ for some $m$ and $A^k \neq I$ for $ 0 < k < m$
(In other words, I want to construct an element $g$ of a finite group that has order $m$ and represent it in $GL(V)$)
If $K = \mathbb{C}$, I can construct $A$ as a diagonal matrix with values from the unity of $\mathbb{C}$, i.e. $\{ z \in \mathbb{C}: z^m = 1 \}$
then, I will get $A^m = I~~~\forall m \in \mathbb{N}$
If $K = \mathbb{R}$, I can construct $A$ as a matrix with block like:
$
\begin{bmatrix}
R_m & 0 \\
0 & I_{n-2} 
\end{bmatrix}
$ where $R_m = \begin{bmatrix}
\cos ( \frac{2 \pi}{m} ) & -\sin( \frac{2 \pi}{m} ) \\
\sin ( \frac{2 \pi}{m} ) & \cos( \frac{2 \pi}{m} ) \\
\end{bmatrix}
$
Then, I can also get $A^m = I~~~\forall m \in \mathbb{N}$
However, if $K = \mathbb{Q}$, the problem is different, I don't think I can construct $A$ for every $m$.
If I think the problem as a cyclic subgroup $C_m$ of symmetric group $S_n$, and then I can represent the element of $C_n$ by permuting the rows of a identity matrix, e.g.
For $n=3$, let $A = \begin{bmatrix}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0 
\end{bmatrix}$, then $A^3 = I$, but $A^k \neq I$ for $0 < k < 3$
I can do the same thing for all $m \leq n$ by fixing $n-m$ rows of the identity matrix. How about if $m > n$?
By adding odd numbers of minus sign to the rows of the identity matrix, like: 
For $n=3, m=6$, let $A = \begin{bmatrix}
0 & 0 & -1\\
1 & 0 & 0\\
0 & 1 & 0 
\end{bmatrix}$, then $A^6 = I$, but $A^k \neq I$ for $0 < k < 6$
I think I can construct up to $m \leq 2n$ by using the above method. (But I think some $m$ is missed, like $m=5$ for $n=3$)
Is it possible to construct $A$ for $m > 2n$? 
(In other words, is there any faithful linear representation of the cyclic group $C_m$, $m > 2n$ on $GL(n,\mathbb{Q})$?)
I know that there is a bound of the order of an element $h$ from $S_n$ by Landau's function $g(n)$.
By permuting the rows of an identity matrix and adding sign, the maximum order I can represent should be bounded by $2g(n)$, right?
For $m$ between $n < m \leq 2g(n)$, are there any positive integer $m$ that I cannot construct $A$ s.t. $A^m = I$ and $A^k \neq I$ for $0 < k < m$? What are they?

REPLY [3 votes]: However, if $K = \mathbb{Q}$, the problem is different, I don't think I can construct $A$ for every $m$.

This is correct.
the following result is (I think) well known
Let $m = p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ be the prime factorization with $p_1<\cdots < p_k$. The group $\mathrm{GL}_n(\mathbb Q)$ has an element of order $m$ if and only if

$\displaystyle \sum_{i=1}^k(p_i-1)p_i^{\alpha_i-1} -1 \leq n$ for $p_1^{\alpha_1} = 2$, or
$\displaystyle \sum_{i=1}^k(p_i-1)p_i^{\alpha_i-1} \leq n$ otherwise.

One proof involves cyclotomic polynomials.
So for instance $\mathrm{GL}_2(\mathbb Q)$ and $\mathrm{GL}_3(\mathbb Q)$ have no element of order $5$, only orders $1,2,3,4$ or $6$.
Several papers cover this topic. For instance 
On the Maximum Order of Torsion Elements in
GL n, Z. and Aut F . or 
The Maximal Orders of Finite Subgroups in GLn(Q)
Shmuel Friedland
Proceedings of the American Mathematical Society
Vol. 125, No. 12 (Dec., 1997), pp. 3519-3526