Torsion subgroup of $SL(n,\mathbb Z)$ Let $G$ be subgroup of $SL(n,Z)$ such that for any $g\in G$ there exists integer $m\geq1$ $g^m=1$.
Show that there exists $N\geq1$ such that for any $g\in G$ , $g^N=1$
I know $m$-th root of unity is the eigenvalue of elements, any element is diagonalizable matrix over complex number but I don't know how to use facts?
any suggestion?
 A: I think I've got it (but check it, it won't be the first time I produce a wrong proof!). 


*

*First step: Let $\mathcal P$ the set of monic polynomials of degree $n$, with coefficients lying in $\mathbb Z$, and the roots in the unit circle of the complex plane. Then $\mathcal P$ is finite. Indeed, fix $0\leq k\leq n-1$ and for $P\in\mathcal P$, $P=X^n+\sum_{j=0}^{n-1}a_jX^j$, with roots $\lambda_1,\ldots,\lambda_n$, we have 
$$a_j=(-1)^j\sum_{J\subset\{1,\ldots,n\},|J|=j}\prod_{k\in J}\lambda_k$$
so $|a_j|\leq \binom nj$ and since $a_j$ is an integer it can take only a finite number of values.

*Second step: For each $g\in G$, the characteristic polynomial of $g$, $p_g$, is an element of $\mathcal P$, so the set of all eigenvalues of the elements of $G$ is finite, and is contained in $\bigcup_{n\geq 1}\mathbb U_n$, where $\mathbb U_n$ is the sets of $n$-th roots of unity. So in fact the eigenvalues are contained in $\bigcup_{j=1}^{n_0}\mathbb U_{k_j}$, where $k_j$ are natural numbers $\geq 1$. Taking $N:=\operatorname{ppcm}(k_j,1\leq j\leq n_0)$, we get the wanted result.

A: The characteristic polynomial of $g\in G$ has degree $n$, and thus the minimal polynomial satisfied by $g$ has degree $\le n$ and divides $x^m-1$, i.e. is a cyclotomic polynomial. The degree of $k^{th}$ cyclotomic polynomial is $\phi(k)$. Thus, if $g$ satisfies $k^{th}$ cyclotomic polynomial then $\phi(k)\le n$. This is a finite set, and the $\operatorname{lcm}$ of this set works.
A: Lemma A2 in the appendix of Swinnerton-Dyer's "Brief guide to Algebraic Number Theory" that might help solving your question. I am really sorry for giving this reference this way. I think this Lemma (or a similar version) has a commonly known name, but I just can not remember it. It states
Let $G$ be a f.g. abelian and torsion free group, and let $H$ be a subgroup. Then there exists a minimal base $x_1, \dots, x_n$ of $G$ and integers $m_1, \dots m_r$, $r≤n$, such that $m_{i+1}$ divides $m_{i}$ and $m_1 x_1, \dots, m_r x_r$ are a bases for $H$.
maybe the condition "abelian" can be withdrawn  
