# Thread: Algebra, Problems For Fun (39)

1. ## Algebra, Problems For Fun (39)

It's a well-known result that in a field $\displaystyle F,$ every finite subgroup of $\displaystyle F^{\times}$ is cyclic. A much less known but more interesting result is this:

Let $\displaystyle D$ be a division ring, i.e. a field which is not necessarily commutative. Prove that every finite abelian subgroup $\displaystyle G$ of $\displaystyle D^{\times}$ is cyclic.

Suggestion:
Spoiler:
Let $\displaystyle k$ be the center of $\displaystyle D$ and $\displaystyle C=\{c_1g_1 + \cdots + c_ng_n: \ \ n \in \mathbb{N}, c_j \in k, \ g_j \in G \}.$ Show that $\displaystyle F=\{xy^{-1}: \ x,y \in C, \ y \neq 0 \} \subseteq D$ is a field and $\displaystyle G \subseteq F^{\times}.$

2. Originally Posted by NonCommAlg
Suggestion:
Spoiler:
...Show that $\displaystyle F=\{xy^{-1}: \ x,y \in C, \ y \neq 0 \} \subseteq D$ is a field...
Why have you included a $\displaystyle y^{-1}$ here? Is it known that $\displaystyle C$ is a field?

3. Originally Posted by Swlabr
Why have you included a $\displaystyle y^{-1}$ here? Is it known that $\displaystyle C$ is a field?
$\displaystyle y^{-1} \in D,$ because $\displaystyle D$ is a division ring. the claim is that $\displaystyle C$ is a field.

4. Originally Posted by NonCommAlg
$\displaystyle y^{-1} \in D,$ because $\displaystyle D$ is a division ring. the claim is that $\displaystyle C$ is a field.
ooooh-I've been getting integral domains mixed up with division rings for a long time...I thought I'd sorted that problem...ah well, this new information should certainly make this question easier!