Prove that the characteristic polynomial of any $g\in G$, where $G$ is a finite subgroup of $GL(n,\mathbb Q)$, has integer coefficients. 
Let $G$ be a finite subgroup of the group of invertible $n\times n$ rational matrices. Show that the characteristic polynomial of any $g\in G$ has integer coefficients.


My attempt:
Suppose the order of group $G$ is $n$, then for every $g\in G\leqslant GL(n,\mathbb Q)$, we have $g^n=I$, the identity matrix. Hence the minimal polynomial $m(x)\in\mathbb Q[x]$ of $g$ divides $x^n-1$. 
If $n=1$ or $n=2$, then by trivial verification, we know that the characteristic polynomial of any $g$ has integer coefficients.
What about other situations? Indeed, we claim that $m(x)|(x^n-1)$ but there are many possibilities for $m(x)$. Any help?
 A: You are using the symbol $n$ for two different numbers, which may lead to some confusion and/or misunderstanding. I'll denote the order of $G$ by $k$ instead.
That the minimial polynomial $m(x)\in\Bbb{Q}[x]$ of $g$ has integer coefficients follows immediately from Gauss's lemma: Because $m(x)$ divides $x^k-1$ we have
$$x^k-1=m(x)q(x),$$
for some $q(x)\in\Bbb{Q}[x]$. Because $x^k-1$ and $m(x)$ are monic, also $q(x)$ is monic and so Gauss's lemma tells us that $m(x),q(x)\in\Bbb{Z}[x]$.

As there are many variations of Gauss's lemma, let me clarify what I mean by Gauss's lemma:

If $g_1,g_2\in\Bbb{Q}[x]$ are monic and such that $g_1g_2\in\Bbb{Z}[x]$, then $g_1,g_2\in\Bbb{Z}[x]$.

The proof is simple, and relies solely on the fact that the product of two primitive polynomials is again primitive. (A polynomial with integer coefficients is called primitive if the greatest common divisor of its coefficients is $1$)
Proof. Let $c_1$ and $c_2$ denote the least common multiples of the denominators of the coefficients of $g_1$ and $g_2$, respectively. Then $c_1g_1,c_2g_2\in\Bbb{Z}[x]$ and both are primitive, and so their product is also primitive, which is
$$(c_1g_1)(c_2g_2)=c_1c_2g_1g_2\in\Bbb{Z}[x],$$
where $g_1g_2\in\Bbb{Z}[x]$. Therefore $c_1c_2=\pm1$, which implies $c_1,c_2=\pm1$ and hence $g_1,g_2\in\Bbb{Z}[x]$.
