Why is the multiplicative group of a finite field cyclic? [duplicate]

Why is the multiplicative group $(K\smallsetminus\{0\},\cdot)$ of a finite field $(K,+,\cdot)$ always cyclic?

• This might help. – Burak Jun 17 '14 at 17:40
• In most cases, if you’re asking the question of why something is true, well, that’s the whole point of mathematics, and you should try to find a proof. If you can’t understand the proof (happens to all of us!), then you might ask here for help with understanding. – Lubin Jun 17 '14 at 17:48
• In general, it sounds better to ask "How do you prove Result X?" rather than ask "Why is Result X true?" They sort of mean the same thing but the first one is much clearer :) – rschwieb Jun 17 '14 at 17:54
• Keith Conrad has collected six proofs of this theorem for the field $\mathbb Z/p$ at math.uconn.edu/~kconrad/blurbs/grouptheory/cyclicFp.pdf. – lhf Jun 17 '14 at 18:40
• To @MartinBrandenburg who marked this as duplicate, I don't think so, for two reasons: 1) I'm asking about the whole group, not finite subgroups, and 2) I'm asking about a finite field, whereas the question this question has been marked as possible duplicate of asks about the subgroups of a generic field's multiplicative group. – MickG Jun 18 '14 at 12:37

3 Answers

In a field, $X^d-1$ can have at most $d$ roots, so there can always be at most $d$ elements whose order divides $d$.

The only commutative finite groups with this property are the cyclic groups.

If $G$ is any other commutative finite group then it has a subgroup of the form $(\Bbb Z/p \Bbb Z)^2$ for some integer $p$, which means it has $p^2$ elements of order dividing $p$.

• That argument depends on the structure of abelian groups. – lhf Jun 17 '14 at 18:17
• Uh-huh. And how do you prove that «the only commutative finite groups with this property are the cyclic groups»? – MickG Jun 17 '14 at 18:33
• @MickG: There is also more elementary reason for this. Firstly oberve that analyzing the structure of a cyclic finite group of order $n$ yields the Euler's formula $\sum_{d \mid n}\varphi(d)=n$, where $\varphi$ is the Euler's totient function. Now, if a finite group $G$ has at most $d$ elements with order dividing $d$ for every $d$ dividing $|G|$, then using Euler's formula one can see that it necessarily has an element of order $|G|$. – Pavel Čoupek Jun 17 '14 at 20:02
• @Pavel Coupek, I am not understanding your argument for existence of an element of order |G|. – Martund Aug 23 '19 at 3:24
• @Martund Let me say it a bit differently: A cyclic group of some order $m$ has precisely $\varphi(k)$ elements of order $k$ for every divisor $k$ of $m$, which also proves the Euler's formula. Now if $g \in G$ is of order $d$, then $\langle g \rangle$ consists of $d$ elements of order dividing $d$, so it contains all such elements of $G$ by assumption. In particular, $\langle g \rangle$ is the unique cyclic subgroup of $G$ of order $d$, and in particular, $G$ has in this case $\varphi(d)$ elements of order $d$. So $G$ has at most $\varphi(d)$ elements of order $d$ (either $0$ or $\varphi(d)$). – Pavel Čoupek Aug 23 '19 at 3:55

Let $$G$$ be the multiplicative group of a finite field, $$n$$ its order. Let $$d$$ be a divisor of $$n$$, $$\psi(d)$$ the number of elements order $$d$$ in $$G$$. Suppose there exists an element $$a$$ of $$G$$ whose order is $$d$$. Let $$H$$ be the subgroup of $$G$$ generated by $$a$$. Then every element of $$H$$ satisfies the equation $$x^d = 1$$. Since the number of the solutions of $$x^d = 1$$ is less than or equal to $$d$$ and the order of $$H$$ is $$d$$, $$H = \{x \in G\mid x^d = 1\}$$. Therefore $$\psi(d) = 0$$ or $$\phi(d)$$, where $$\phi(d)$$ is the Euler's function, i.e. the number of elements of order $$d$$ of a cyclic group of order $$d$$. Since $$\sum_{d\mid n} \psi(d) = n = \sum_{d\mid n} \phi(d)$$, $$\psi(d) = \phi(d)$$ for all $$d\mid n$$. In particular $$\psi(n) = \phi(n)$$, which means there exists an element of order $$n$$ in $$G$$. This completes the proof.

Let $G$ be the multiplicative group of a finite field, $n$ its order. Let $x$ be an element of $G$, $d$ its order. Let $H$ be the subgroup of $G$ generated by $x$. If $G = H$, we are done. Suppose otherwise. There exists $y \in G - H$. Let $m$ be the order of $y$, $l$ = lcm$(d, m)$. Suppose $d = l$. Then $m\mid d$. Hence $y^d = 1$. This contradicts the fact that the number of solutions of the equation $X^d = 1$ is less than or equal to $d$. Hence $d \lt l$. By the answer to this question, there exists an element $z$ of order $l$ in $G$. We can repeat this process until we find a generator of $G$. This completes the proof.