## Finite subgroups of GL(n, Z)

Posted: February 17, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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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 $\text{GL}(n,\mathbb{Z}).$