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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.

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    Newman is wrong.2018-02-12

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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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    @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