# Units of finite rings

1. Dec 12, 2008

### mooshasta

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 $$a^q = 1$$.

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 $$a^{\phi (m)} = 1 (mod \, m)$$, 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?

Thanks

2. Dec 12, 2008

### sutupidmath

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.

3. Dec 13, 2008

### mooshasta

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..