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Math Help - Algebra, Problems For Fun (39)

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

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

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


    Suggestion:
    Spoiler:
    Let k be the center of D and C=\{c_1g_1 + \cdots + c_ng_n: \ \ n \in \mathbb{N}, c_j \in k, \ g_j \in G \}. Show that F=\{xy^{-1}: \ x,y \in C, \ y \neq 0 \} \subseteq D is a field and G \subseteq F^{\times}.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    Suggestion:
    Spoiler:
    ...Show that F=\{xy^{-1}: \ x,y \in C, \ y \neq 0 \} \subseteq D is a field...
    Why have you included a y^{-1} here? Is it known that C is a field?
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    Why have you included a y^{-1} here? Is it known that C is a field?
    y^{-1} \in D, because D is a division ring. the claim is that C is a field.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    y^{-1} \in D, because D is a division ring. the claim is that 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!
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