1
$\begingroup$

Consider the equivalence relation $m\sim n$ defined to be $\frac {m-n}p=z$ (i.e., when $m-n$ is divisible by p) where $m,n,p,z\in \mathbb{Z}$ and $p$ is prime. Now suppose that $a$ is some integer not divisible by $p$. Prove that $[a]^{p-1}=[1]$ where [1] denotes the equivalence class of 1. Another common way to write this is $a^{p-1} \equiv 1$ $\ $(mod $p$).

I understand that there is some intent to prove multiplicative inverses here, but I'm a bit confused as to how one would approach it.

  • 3
    hint: In a group $a^{|G|} = e$ $\ $ for every a in G where $|G|$ is the order of the group and $e$ is the identity.2012-09-28
  • 3
    My favorite proof is http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem#Second_method2012-09-28
  • 2
    This was never asked before?2012-11-08
  • 0
    @hardmath: And note that this is a very rare answer by B.D. that actually uses italics for variables (at least in the quote).2013-02-19
  • 0
    @MarcvanLeeuwen: The new duplicate question process seems to have a side effect of removing a comment (pointing to an earlier Q) when the banner is added. In a couple of cases, as here, my comment was slightly more substantive than just a link to the dup, so I'm not entirely happy with the change.2013-02-19

1 Answers 1