Question

Prove that the set of all the non zero elements in a field is a multiplicative group. Use Lagrange's Theorem to prove that in a finite field of m elements x^m = x for every x.

Answer 1

Question 1.

Prove that the set of all the non-zero elements in a field is a multiplicative group. Use Lagrange’s Theorem to prove that in a finite field of $m$ elements $x^{m}=x$ for every $x$.

Solution

Let $\mathbb{F}$ be a field and $\mathbb{F}^{*}$ denote the set of all non-zero elements of $\mathbb{F}$. Prove that $\mathbb{F}$ is a multiplicative group. Since $1$ is assumed to be distinct from , then $1\in\mathbb{F}^{*}$ and it is the identity of $\mathbb{F}^{*}$, because

$1\cdot a=a\cdot 1=a$

for any $a\in\mathbb{F}$. Furthermore, if $a\neq 0$ and $b\neq 0$, then $ab\neq 0$ since otherwise we could multiply $ab$ on $b^{-1}$ (it exists, because $b\neq 0$) and get $a=0$, which is not true. So, $\mathbb{F}$ is closed under multiplication. Finally, by one of the field axioms, any non-zero $a$ has the multiplicative inverse, i. e. an element $a^{-1}$, satisfying

$aa^{-1}=a^{-1}a=1.$

This means that $\mathbb{F}$ is closed under taking multiplicative inverses. Thus, $\mathbb{F}$ is a (commutative) group under multiplication.

If $\mathbb{F}$ consists of $m$ elements, then $\mathbb{F}^{*}$ has the order $m-1$, because $\mathbb{F}^{*}=\mathbb{F}\setminus\{0\}$. By Lagrange theorem, the order of any element $a\in\mathbb{F}^{*}$ divides the order of $\mathbb{F}^{*}$, which is $m-1$. Therefore, $a^{m-1}=1$ for any $a\in\mathbb{F}^{*}$. Multiplying both sides by $a$, we obtain $a^{m}=a$ for all $a\in\mathbb{F}^{*}$. But this is also true for $a=0$, so $a^{m}=a$ for any $a\in\mathbb{F}$. ∎