# Show that $x^{n-1} =1$ for all non-zero element in a field

Question:

Show that $x^{n-1} =1$ for all non-zero element in a field

Let F be a finite field of order n. Show that $x^{n-1}=1$ for all non-zero $x \in F$.

We have $\left | F \right |=n.$

Recall that a field is an integral domain in which every non-zero element is a unit.

Recall that a unit in a ring is an element in the ring with a multiplicative inverse in R.

Recall that an integral domain is a commutative ring with unity containing no non-zero divisor.

Suppose $\left [ x\neq 0 \right ]\in F$, then, by definition of a field F, $\exists y \in F$ s.t $x\cdot y=1$

Note that $x\cdot y=1 \in F$.

Can someone drop a hint to take me further? Thanks in advance.

• Do you know Lagrange's theorem ? – Captain Lama May 11 '16 at 12:35
• Yes. Is that your hint? – Mathematicing May 11 '16 at 12:36
• I do not. @Joanpemo – Mathematicing May 11 '16 at 12:38
• @Mathematicing Yes, I just realized back then we don't need that here. Read my answer. – DonAntonio May 11 '16 at 12:39

## 2 Answers

$\;F\;$ is a field $\;\implies F^*=F\setminus\{0\}\;$ is a multiplicative group of order

$$\;|F|-1=n-1\;\implies\;\forall\,x\in F^*\;,\;\;x^{n-1}=1$$

since in a finite group it is always true that the order of any element divides the order of the group (as commented, this is Lagrange's Theorem)

• The n-1 is a bit hard to figure out why. – Mathematicing May 11 '16 at 13:24
• @Mathematicing Really? You said $\;|F|=n\;$ , so then $\;|F^*|=|F\setminus\{0\}|=n-1\;$ ...I think it is trivial. – DonAntonio May 11 '16 at 15:04
• I noted the poster's comment above and realised. – Mathematicing May 12 '16 at 1:30

$F\setminus \{0\}$ is a finite group with identity element $1$, which has ...(?) elements. Now use Lagrange's theorem.

• I hope this question isn't silly: When you say F is a group, is it a group by virtual of the fact that if forms an Abelian group under addition? – Mathematicing May 11 '16 at 12:43
• @Mathematicing - when people refer to $F \setminus \{0\}$, it is usually to the multiplicative abelian group formed by $\times$ with identity $1$ – Siddharth Bhat May 11 '16 at 12:44
• @Mathematicing ....under multiplication – user 1 May 11 '16 at 12:44
• see here. it has: "In mathematics, a field is one of the fundamental algebraic structures used in abstract algebra. It is a nonzero commutative division ring, or equivalently a ring whose nonzero elements form an abelian group under multiplication" – user 1 May 11 '16 at 12:47
• in the link above: "...a ring whose nonzero elements form an abelian group under multiplication" – user 1 May 11 '16 at 13:31