Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ a finite field of characteristic $p$. Show that $p-1$ divides $|F|-1$. (We shall see later that $|F|$ is a power of $p$.)

I am able to solve this by first showing $|F|$ is a power of $p$. If $q$ divides $|F|$ for another prime $q$, then by Cauchy's theorem some element $x$ has order $q$ in the additive group, so $qx=0$. But $qx=px=0$, so the order of $x$ has order $\gcd(p,q)=1$, a contradiction. Thus $|F|=p^n$ for some $n$, and $p-1$ divides $|F|-1$ since $p^n-1=(p-1)(p^{n-1}+\cdots+1)$.

The remark at the end of the exercise that we shall later see $|F|$ is a prime power suggests to me that there may be an alternative proof that doesn't resort to this fact. Is there some obvious observation I'm overlooking, because the proof I believe I have seems simple enough.

share|cite|improve this question

A simpler way to prove that the order of $F$ is a power of $p$ is to note that $F$ is a vector space over its prime field $\mathbb{F}_p$. Thus, as an abelian group, it is isomorphic to a direct sum of copies of $\mathbb{Z}/p\mathbb{Z}$, hence has order $p^n$ for some $n\gt 0$.

To show the desired result without showing that $|F|$ is a power of $p$, note that the group of units of $F$ contains the group of units of $\mathbb{F}_p$, which has order $p-1$. By Lagrange's Theorem, $p-1$ must divide $|F|-1$.

share|cite|improve this answer
Ah right, thanks! – Adelaide Dokras Jun 17 '12 at 1:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.