There are 73 conjugacy classes of finite subgroups in $\operatorname{GL}(3,\mathbb{Z})$. If you take 73 representatives, you will find group-subgroup relations between them. There must exist an overview from the maximal subgroups (I believe there are 14?) to the trivial group, but I don't seem to be able to find it.
Subgroup relations in $GL(3,\mathbb Z)$
3
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group-theory
finite-groups
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0According to Morris Newman, in his book "Integral matrices",page 181, there are 70, not 73, conjugacy classes of finite subgroups in $\mathrm{GL}(3,\mathbb Z)$. Where did you get 73? – 2011-10-10
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0Newman is wrong. – 2018-02-12
1 Answers
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There is a natural homomorphism from $\operatorname{GL}(3,\mathbb{Z})$ to $\operatorname{GL}(3,\mathbb{F}_3)$, by which each finite subgroup of $\operatorname{GL}(3,\mathbb{Z})$ is mapped injectively to a subgroup of $\operatorname{GL}(3,\mathbb{F}_3)$. From subgroup lattice in later, one can obtain subgroup lattice in $\operatorname{GL}(3,\mathbb{Z})$.
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0Do you mean something like [this](http://tinyurl.com/6598g69)? This is for $GL(2,\mathbb{F}_3)$ so then I would expect 13 groups... I'm obviously missing something. – 2011-09-15
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0@Wox: GL(n,Z/mZ) has more (conjugacy classes of) finite subgroups than GL(n,Z), but if you only look at the subgroups of GL(n,Z/mZ) that lift to subgroups of GL(n,Z) then you should get something closer to what you are looking for. – 2011-09-25