## Finite subgroups of the multiplicative group of a division ring

Posted: September 20, 2011 in Division Rings, Noncommutative Ring Theory Notes
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It is a well-known fact that a finite subgroup of the multiplicative group of a field is cyclic. We will prove this result shortly.  We will also extend it to any division ring of non-zero characteristic. Note that this result is not necessarily true in a division ring of zero characteristic. For example, in the division ring of real quaternions, the subgroup $\{\pm 1, \pm i, \pm j, \pm k \}$ is not even abelian let alone cyclic. We will also prove that finite abelian subgroups of the multiplicative group of any division ring are cyclic.

Theorem. Every finite subgroup of the multiplicative group of a field is cyclic.

Proof. Let $F$ be a field and let $G$ be a finite subgroup of $F^{\times}.$ Let $|G|=n$ and, for any divisor $d$ of $n,$ let $f(d)$ be the number of elements of $G$ of order $d.$ Obviously

$\sum_{d \mid n} f(d) = n. \ \ \ \ \ \ (1)$

Let $\varphi$ be the Euler’s totient function. Recall from number theory that

$\sum_{d \mid n} \varphi(d)=n. \ \ \ \ \ \ (2)$

Claim. If $d \mid n$ and $f(d) \neq 0,$ then $f(d)=\varphi(d).$

Proof of the claim. Since $f(d) \neq 0,$ there exists $g \in G$ such that $o(g)=d.$ Let $H = \langle g \rangle$ and $p(x)=x^d - 1.$ Then every element of $H$ is a root of $p(x).$ But $p(x)$ has at most $d$ roots in $F.$ Thus $H$ is exactly the set of roots of $p(x).$ Finally, the fact that an element $g^m \in H$ has order $d$ if and only if $\gcd(m,d)=1$ implies $f(d)=\varphi(d). \ \Box$

It now follows from $(1), \ (2)$ and the claim that $f(d)=\varphi(d)$ for all $d \mid n.$ In particular, $f(n)=\varphi(n) \neq 0$ and so $o(g)= n = |G|$ for some $g \in G. \ \Box$

Corollary 1. Every finite abelian subgroup of the multiplicative group of a division ring is cyclic.

Proof. Let $D$ be a division ring with the center $Z.$ Let $G$ be a finite abelian subgroup of $D^{\times}$ and put $F=\sum_{g \in G} Zg.$ It is obvious that $F$ is a commutative domain and $G \subset F.$ Also,  since $G$ is finite, $F$ is a finite dimensional vector space over $Z$ and thus every element of $F$ is algebraic over $Z.$ Let $0 \neq c \in F$ and suppose that $q(x)=x^m + \ldots + a_1x + a_0 \in Z[x]$ is the minimal polynomial of $c$ over $Z.$ Then $a_0 \neq 0$ and so $c(c^{m-1} + \ldots + a_1)(-a_0^{-1})=1.$ Therefore $F$ is a field and we are done by the above theorem. $\Box$

Corollary 2.  Every finite subgroup of the multiplicative group of a division ring of non-zero characteristic is cyclic.

Proof. Let $D$ be a division ring with $\text{char}(D)=p > 0.$ Let $\mathbb{F}_p$ be the prime subfield of $D$ and suppose that $G$ is a finite subgroup of $D^{\times}.$ Let $F = \sum_{g \in G} \mathbb{F}_p g.$ Clearly $F$ is a finite subring of $D$ and $F$ contains $G.$ Let $0 \neq c \in F.$ Since $D$ is a domain, $F$ is a domain too. Thus $\{cx : \ x \in F\}=F$ and so $cx = 1$ for some $x \in F.$ Therefore $F$ is a division ring. But, by Wedderburn’s little theorem, a finite division ring is a field. So $F$ is a field and we are done by the above theorem. $\Box$