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Jul 8, 2013 at 9:02 history closed Ian Agol
Theo Johnson-Freyd
Andrey Rekalo
Carlo Beenakker
user10534
Duplicate of Number of conjugacy classes in GL(n,Z)
Jul 8, 2013 at 6:33 comment added JHM @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?
Jul 8, 2013 at 4:08 comment added Venkataramana @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.
Jul 8, 2013 at 3:15 review Close votes
Jul 8, 2013 at 9:06
Jul 8, 2013 at 3:01 comment added JHM @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.
Jul 8, 2013 at 2:51 comment added Venkataramana conjugacy class of torsion elements should be easy: they look like companion matrices of products $f$ of cyclotomic polynomials, with no repeated roots.
Jul 8, 2013 at 1:46 answer added Igor Rivin timeline score: 2
Jul 8, 2013 at 0:04 review First posts
Jul 8, 2013 at 0:28
Jul 7, 2013 at 23:47 history asked Incognito CC BY-SA 3.0