Finite subgroups of GL(n, Z)

Posted: February 17, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

Definition. Let R be a commutative ring with unity and let n be a positive integer. Let M_n(R) be the set of n \times n matrices with entries in R. We define \text{GL}(n,R) to be the set of units of M_n(R). Clearly \text{GL}(n,R) is a group under matrix multiplication.

Problem. (Minkowsky) Prove that for any integer n \geq 1, the number of finite subgroups of \text{GL}(n,\mathbb{Z}) is finite.

Solution. Let \mathbb{F}_3 be the field of order 3. In order to prove the theorem, we only need to prove that every finite subgroup of \text{GL}(n,\mathbb{Z}) is isomorphic to some subgroup of \text{GL}(n, \mathbb{F}_3), because \text{GL}(n, \mathbb{F}_3) is a finite group. So suppose that G is a finite subgroup of \text{GL}(n,\mathbb{Z}). Define the map

\varphi : \text{GL}(n,\mathbb{Z}) \longrightarrow \text{GL}(n, \mathbb{F}_3)

by \varphi(A)=\overline{A}, where the entries of \overline{A} are defined to be the entries of A modulo 3. Obviously \varphi is a group homomorphism. Let I be the identity matrix in \text{GL}(n,\mathbb{Z}). We consider two cases.

Case 1 . G \cap \ker \varphi = \{I\}. In this case G \cong \varphi(G) and we are done.

Case 2 . G \cap \ker \varphi \neq \{I\}. Let K = G \cap \ker \varphi. Let p be a prime divisor of |K|. By Cauchy, there exists an element A \in K such that

A \neq I \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

and

A^p=I. \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

Since A \in K \subseteq \ker \varphi, we must have \overline{A}=\varphi(A)=\overline{I}. Thus, if A =[a_{ij}], then a_{ii} \equiv 1 \mod 3, for all 1 \leq i \leq n, and a_{ij} \equiv 0 \mod 3, for all i \neq j. Therefore

A = I + 3B, \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)

for some B \in M_n(\mathbb{Z}). If B=0, then A=I, contradicting (1). So B \neq 0. Let d be the greatest common divisor of the non-zero entries of B. So I can write B = dC, where 0 \neq C \in M_n(\mathbb{Z}) and the greatest common divisor of the non-zero entries of C is one. Now, (3) becomes A=I + 3dC and so, by (2),

\displaystyle I = A^p=(I + 3dC)^p = I + 3pdC +\sum_{i=2}^p \binom{p}{i}(3d)^iC^i.

Thus

\displaystyle pC = -3d \sum_{i=2}^p \binom{p}{i}(3d)^{i-2}C^i. \ \ \ \ \ \ \ \ \ \ (4)

Therefore p=3 and d=1 because the greatest common divisor of the entries of C is one. So (4) becomes C = -\sum_{i=2}^3 \binom{3}{i}3^{i-2}C^i = -3(C^2+C^3), which is absurd because then 3 would divide every entry of C. This contradiction proves that basically case 2 can never happen. \Box

Remark. We proved that every finite subgroup of \text{GL}(n,\mathbb{Z}) is isomorphic to a subgroup of \text{GL}(n,\mathbb{F}_3). A similar argument shows that if q>2 is any prime number, then every finite subgroup of \text{GL}(n,\mathbb{Z}) is isomorphic to a subgroup of \text{GL}(n,\mathbb{F}_q).

In the next post I will show that every finite group of order n is isomorphic to a subgroup of \text{GL}(n,\mathbb{Z}).

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Comments
  1. Chandrasekhar says:

    Hi–

    May I know in what topics did you encounter such theorems. IS it representation theory. Because i have also learnt Basics in abstract algebra but never encountered these things.

    • Yaghoub says:

      Well, the reason that I put this post in the elementary section is that the idea of the proof is elementary not because it can be found in elementary textbooks. 🙂

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