In a finite field $F$, how is $a^{|F|}=a$?

Question

Let $F$ be a finite field with $|F|=q$. If $a\in F$, I have to prove $$a^q=a$$.

We know that if $a^r=1$, then $r|q$ (treating $F$ is a group under multiplication, and using Lagrange's theorem). Hence, $a^q=a^{r\frac{q}{r}}=1^{\frac{q}{r}}=1$. This is because $\frac{q}{r}$ is an integer.

How is it possible then that $a^q=a$?

Thanks in advance!

Answer 1

Hint: If $a=0$ this is obvious. If not, then you can say $a^{|F^\times|}=1$ (why?), and what is $|F^\times|$?

Answer 2

If $F$ is a field, then its units form a group under multiplication. There are $|F|-1$ units in $F$, and therefore by Lagrange's theorem, $a^{|F|-1|} = 1$, which is the identity element of the the aforementioned group. Multiplying both sides by $a$ yield the result you want, as well as allow $0$ to be part of the equation.