Finite Subgroups of general linear group If $H$ is a finite subgroup of $GL(n,\mathbb{Z})$ then by Minkowskie's theorem, it injects to a subgroup of $GL(n,\mathbb{Z}/p\mathbb{Z})$ under the natural map from $GL(n,\mathbb{Z})$ to $GL(n,\mathbb{Z}/p\mathbb{Z})$, where $p$ is an odd prime.
What are restrictions on subgroups $H$ that can inject to a subgroup of $GL(n,\mathbb{Z}/2\mathbb{Z})$ under the natural map from $GL(n,\mathbb{Z})$ to $GL(n,\mathbb{Z}/2\mathbb{Z})$
 A: Let me make some isolated remarks.  Let $K(n,l)$ be the kernel of the map $GL(n,\mathbb{Z}) \rightarrow GL(n,\mathbb{Z}/l\mathbb{Z})$.  


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*The proof of the fact you alluded to above also proves that all finite subgroups of $GL(n,\mathbb{Z})$ inject into $GL(n,\mathbb{Z}/4\mathbb{Z})$.  What you want to understand, therefore, are the intersections of finite subgroups of $GL(n,\mathbb{Z})$ with the kernel of the map $K(n,2) \rightarrow GL(n,\mathbb{Z}/4\mathbb{Z})$.

*The next observation is that that all torsion elements of $K(n,2)$ are 2-torsion.  I don't have time to write out a proof of this, but it is a fairly easy exercise.  Hint : prove that if $M \in K(n,2)$, then $M^2 \in K(n,4)$.

*Finally, a classical theorem in representation theory says that, with an appropriate choice of integral basis, any $2$-torsion element of $GL(n,\mathbb{Z})$ is a direct sum of copies of the identity, multiplication by $-1$, and $\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array} \right)$.  For instance, this is Theorem 74.3 in Curtis-Reiner's book "Representation Theory of Finite Groups and Associative Algebras".  In $K(n,2)$, the last factor cannot occur.  We conclude that every torsion element of $K(n,2)$ can, with respect to an appropriate choice of integral basis, be written as a diagonal matrix with $1$'s and $-1$'s on the diagonal.
