Today, we are going to prove that if $F$ is a finite field, the the multiplicative group $F^\times$ of $F$ is a cyclic group. Let’s start with an example of this phenomenon. Take $\F_5 = \{0,1,2,3,4\}$. Now, if we take powers of $2$ modulo $5$ we get

$$2,2^2=4,2^3 = 3, 2^4 = 1.$$ Therefore, we see that $2$ generates $\F_5^\times$. In other words, $\F_5^\times\cong \Z/4$. In general, we have

$$F_{p^n}^\times\cong\Z/(p^n-1).$$

## Cyclic groups

To start our journey, let us go back to some fundamentals: cyclic groups. Despite their seemingly simple nature, cyclic groups still have some interesting properties. Let $n\geq 1$ be an integer, and consider the cyclic group $\Z/n$. Suppose $d$ divides $n$, or in notation: $d\mid n$. How many elements of order $d$ are there in $\Z/n$? One such element is $n/d$. If $k$ is another such element, then $dk = mn$ for some $m$ so that we see all elements in $\Z/n$ of order $d$ are a multiple of $n/d$. I’ll leave it to the reader to verify all elements are of the form $an/d$ where $a$ is relatively prime to $d$. (Hint: if $a$ is not relatively prime to $d$ then there exists a $d’$ smaller than $d$ such that $d’an/d$ is zero in $\Z/n$).

Thus, the number of elements of order $d$ in $\Z/n$ is $\varphi(d)$, where $\varphi(d)$ denotes the number of natural numbers relatively prime to $d$. The function $\varphi$ on natural numbers is called Euler’s totient function. So actually, the number of elements of order $d$ in $\Z/n$ does not depend on $n$.

If we introduce the notation $\psi_G(d)$ to denote the number of elements in a group $G$ of order $d$ then we have shown that

$$\psi_{\Z/n}(d) = \varphi(d).$$

## A sufficient condition for cyclicity

Now we consider a very interesting condition for a finite group $G$ of order $n$ to be cyclic.

**Theorem**. Let $G$ be a finite group of order $n$. If for every $d\mid n$, we have

$$|\{ x\in G : x^d = 1\}| \leq d$$then $G$ is a cyclic group.

*Proof.*The assumption is that

$$|\{ x\in G : x^d = 1\}| \leq d$$for every $d\mid n$. We claim that the number of elements in $G$ of order $d$, which we have denoted by $\psi_G(d)$, satisfies the equation

$$\psi_G(d) \leq \varphi(d)$$ for every $d\mid n$. This is true if $\psi_G(d) = 0$, so let us assume that $\psi_G(d) \geq 1$. Then there exists an element $g\in G$ of order $d$. Let $H$ be the cyclic subgroup of $G$ generated by $g$. Since every element $x\in H$ satisfies $x^d = 1$ and $H$ has $d$ elements, we conclude from the assumption that $H$ actually contains every element in $G$ of order $d$.

Because $H\cong \Z/d$, and because $H$ contains every element of $G$ of order $d$, we have

$$\psi_G(d) = \psi_H(d) = \psi_{\Z/d}(d) = \varphi(d)$$ with the last equality being what we proved in the first section of this post.

Notice that we have not yet proved in general that $\psi_G(d) = \varphi(d)$, only that this holds *when we assume the condition* $\psi_G(d) \not= 0$.

Now, every element of $G$ has order $d$ for some $d\mid n$. Therefore,

$$n = \sum_{d\mid n}\psi_G(d)\leq \sum_{d\mid n}\varphi(d) = \sum_{d\mid n}\psi_{\Z/n}(d) = n.$$ This equation proves *unconditionally* that $\psi_G(d) = \varphi(d)$ for all $d\mid n$, and so in particular $\psi_G(n) = \varphi(n)\geq 1$. Hence $G$ contains an element of order $n$ and so $G$ is cyclic.

## The multiplicative group of a finite field

If $F$ is a finite field, then the condition

$$|\{ x\in F^\times : x^d = 1\}| \leq d$$ is satisfied because $F$ is a field so $x^d – 1$ has at most $d$ solutions. Therefore, $F^\times$ is a cyclic group. Notice that we have also proved that any finite subgroup of $F^\times$ is a cyclic group for any field $F$.

(*Want to take a break from algebra? Watch one of my short nature films on Youtube.*)