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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?

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    If $g$ was diagonal, what would $g^m$ look like?2012-04-19
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    @steve both g and $g^m$ are diagonal2012-04-19
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    OK... a better question: what can you say about the minimal polynomial of such a $g$?2012-04-19
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    $x^m-1$ is divide by minimal polynomial2012-04-19
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    The irreducible factors of $x^m-1$ are the cyclotomic polynomials $\Phi_d(x)$ with $d|m$. So the minimal polynomial of $g$ is a product of some of these $\Phi_d(x)$. If $g$ has order $m$, then the least common multiple of the $d$ arising in this factorization must be $m$. That should be enough to enable you to bound $m$ as a function of the degree of the minimal polynomial of $g$.2012-04-19
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    @DerekHolt why should be enough to enable you to bound m as a function of the degree of the minimal polynomial of g?2012-04-20
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    For each $p^d$ in the factorization of $m$ into prime powers, one of the factors of the minimal polynomial of $g$ must have degree a multiple of (and hence at least) $\phi(p^d) = (p-1)p^{d-1}$.2012-04-20

3 Answers 3

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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.
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    @Marc Thanks for editing.2012-04-23
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    I understand the first step. I don't think I understand the second step. Just because $\lambda^{N} =1$ for all eigenvalues $\lambda$ (of every $g\in G$), why does it imply that $g^{N} = 1$ for all $g\in G$?2017-03-16
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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.

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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

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    I don't think the group $G$ of the question is abelian.2012-04-19
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    yes,I am agree with Marc, it is not abelian2012-04-19
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    No, the condition "abelian" cannot be removed from the theorem. For example, the free group of rank $2$ is finitely generated and torsion free. The commutator subgroup, however, is not finitely generated, so you cannot possibly find a basis (free generating set) for $G$ and integers as given to yield a finite generating set, let alone a free generating set, for $H$.2012-04-19