# Let F be a finite field with n elements. Prove $x^{n-1}=1$ for all nonzero x in F.

Let F be a finite field with n elements. Prove that $x^{n-1}=1$ for all nonzero x in F.

I'm not understanding where this proof is going. So we have a ring $Z_{7}$, and we know there are 7 elements in $Z_7$ which are $\{0,1,2,3,4,5,6\}$ (a complete system of residues modulo 7), obviously every nonzero element to the power of 6, (from n-1 where n=7) is congruent to 1 modulo 7. However, I'm not understanding what they want me to show. I could show this by induction but I get the feeling that there is an easier way to show this?

• $F\setminus \{0\}$ is a multiplicative group. Use Lagrange's theorem. – Ragib Zaman May 6 '12 at 7:28
• And you can't assume that $F$ is of the form $\Bbb Z_n$. – Brian M. Scott May 6 '12 at 7:29
• Just out of curiosity, how would you show it by induction? – Tara B May 6 '12 at 10:06

## 3 Answers

As noted by Ragib if $F$ is field then $F - \{0\}$ is a multiplicative group of order $n-1$. Therefore the order of any element $x$ in here must be a divisor of $n-1$, viz. if $m$ is the order of $x$, then $m | n-1$. Hence $qm = (n-1)$ for some $q \in \Bbb{N}$. It follows that

$$x^{n-1} = x^{mq} = (x^m)^q = 1^q = 1.$$

Q.E.D.

A simple consequence of Lagrange's theorem is that for any finite group $G$:

$g^{|G|} = e$

for all $g\in G$.

In your case you have a multiplicative group of order $n-1$. Using the result we must have that $x^{n-1} = 1$ for all non-zero $x\in F$.

The number of answers using Langrange's theorem is too damn high ! ;-)

So I will proceed differently. Let $x\in F$, $x\not= 0$. Then as $F$ is a field, the map $\alpha : y\mapsto x y$ is a bijection from $F^{\times}$ to $F^{\times}$. Therefore we have that $\Pi_{y\in F^{\times}} y = \Pi_{y\in F^{\times}} \alpha(y) = \Pi_{y\in F^{\times}} (xy) = x^{\textrm{Card}(F^{\times})} \Pi_{y\in F^{\times}} y$. Now, the element $\Pi_{y\in F^{\times}} y$ is non-zero because $F$ is a field, so that we get by simplification by $\Pi_{y\in F^{\times}} y$ that $x^{\textrm{Card}(F^{\times})} = 1$. Now, if $n$ is the cardinal of $F$, as $n - 1 = \textrm{Card}(F^{\times})$ because $F$ is a field, we get that $x^{n-1} = 1$, as required.

• This works in every finite abelian group. – Martin Brandenburg Oct 31 '14 at 22:01
• Yes, I know, but no need to mention it, shush ! ;-) – ujsgeyrr1f0d0d0r0h1h0j0j_juj Oct 31 '14 at 22:02