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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.

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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 $|F|-1$.

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Ah right, thanks! –  Adelaide Dokras Jun 17 '12 at 1:55
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