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Units of finite rings

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Let q be the number of units in finite ring R. Show that for all a in R, if a is a unit in R then [tex]a^q = 1[/tex].

    Is there a way to solve this without using group theory? All I can seem to find information on is when a and m are relatively prime then [tex]a^{\phi (m)} = 1 (mod \, m)[/tex], which I'd like to prove using the problem I can't solve.

    2. Relevant equations

    3. The attempt at a solution

    I really haven't been able to get anywhere on this. Are there certain patterns that finite rings always follow, that I can exploit?

  2. jcsd
  3. Dec 12, 2008 #2
    If q is the number of invertibles in the ring R, then it means that the order of V is q. Where V would be the group of invertibles in that ring. That is we know that V( the set of all invertibles in a ring is a group in itself). Now it is clear that if a is invertible(unit) then it belongs to V. So, by lagrange theorem we have the desired result that a^q=1. where 1 is the unity of the ring R.
  4. Dec 13, 2008 #3
    Thank you for your help.

    I'm trying to solve this without using groups or Lagrange's theorem (we didn't learn any of that in class).

    I made progress but I still am a little hung up at one (probably extremely trivial) detail.

    Let S be the set of units {x_1,...,x_n} and let a be a unit of the ring. T = {ax_1,...,ax_n} are all different and therefore T = S, which implies the products (call it z) are the same. a^n*z = z implies a^n = 1.

    I am not sure how to prove that all members of T = {ax_1,...,ax_n} are all unique, although I think it's pretty intuitive..
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