Definition. Let be a commutative ring with unity and let be a positive integer. Let be the set of matrices with entries in We define to be the set of units of Clearly is a group under matrix multiplication.
Problem. (Minkowsky) Prove that for any integer the number of finite subgroups of is finite.
Solution. Let be the field of order In order to prove the theorem, we only need to prove that every finite subgroup of is isomorphic to some subgroup of because is a finite group. So suppose that is a finite subgroup of Define the map
by where the entries of are defined to be the entries of modulo Obviously is a group homomorphism. Let be the identity matrix in We consider two cases.
Case 1 . In this case and we are done.
Case 2 . Let Let be a prime divisor of By Cauchy, there exists an element such that
Since we must have Thus, if then for all and for all Therefore
for some If then contradicting (1). So Let be the greatest common divisor of the non-zero entries of So I can write where and the greatest common divisor of the non-zero entries of is one. Now, (3) becomes and so, by (2),
Therefore and because the greatest common divisor of the entries of is one. So (4) becomes which is absurd because then 3 would divide every entry of This contradiction proves that basically case 2 can never happen.
Remark. We proved that every finite subgroup of is isomorphic to a subgroup of A similar argument shows that if is any prime number, then every finite subgroup of is isomorphic to a subgroup of
In the next post I will show that every finite group of order is isomorphic to a subgroup of