# Conjugacy of torsion subgroups in Gl(n, Z) for small n [duplicate]

Have the conjugacy classes of the torsion subgroups of Gl(n, Z) been determined for small n (say, n<=6)? In general, can much be said about the torsion subgroup?

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## marked as duplicate by Ian Agol, Theo Johnson-Freyd, Andrey Rekalo, Carlo Beenakker, Derek Holt Jul 8 '13 at 9:02

conjugacy class of torsion elements should be easy: they look like companion matrices of products $f$ of cyclotomic polynomials, with no repeated roots. – Venkataramana Jul 8 '13 at 2:51
@Aakumadala: I think you are describing torsion elements in,say, $GL(3, \mathbb{C})$ -- which is a different question than torsion in $GL(n,\mathbb{Z})$. E.g. one of these groups has elements of arbitrarily large finite order, while the other doesn't. – J. Martel Jul 8 '13 at 3:01
@Martel: No, for example, a product $\prod \Phi _d$ of cyclotomic polynomials of total degree $3$, has very few solutions. Surely these give conjugacy classes. The issue would be rational conjugacy vs integral conjugacy, but this is not insoluble. – Venkataramana Jul 8 '13 at 4:08
@Aakumadula: evidently,i see that if $A \in GL(n,Q)$ has order $k$, then the minimal polynomial $m_A$ of $A$ divides $\prod_{d|k} \Phi_d$. Over $\mathbb{C}$ the elementary divisors $d_1|\ldots|d_r=m_A$ then determine $A$ up to conjugacy in $GL(n,C)$. In your initial comment, were you suggesting something of how to determine the conjugacy type of $A$ in $GL(n, Q)$ relative to its elementary divisors? – J. Martel Jul 8 '13 at 6:33

Although every finite subgroup of ${\rm GL}_n({\mathbb Q})$ is conjugate to a subgroup of ${\rm GL}_n({\mathbb Z})$, there are more conjugacy classes in the integral group than in the rational group, and it is considerably easier to compute them over the rationals, which is why they have to up to higher dimensions. – Derek Holt Jul 8 '13 at 9:00