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

Martin Brandenburg
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MickG
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    This might help. – Burak Jun 17 '14 at 17:40
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    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
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    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
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    Keith Conrad has collected six proofs of this theorem for the field $\mathbb Z/p$ at http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cyclicFp.pdf. – lhf Jun 17 '14 at 18:40
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    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
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    @lhf the link now has 7 proofs, not 6, and the URL has changed to http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/cyclicmodp.pdf. – KCd Jun 21 '18 at 18:53
  • I have to prove this statement using the fact that any Abelian group is a module over $\mathbb{Z}$. Can anyone help me with this? – MarlonButBetter Dec 11 '23 at 14:05

3 Answers3

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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.

joriki
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    You say that the number of solutions for $x^d=1$ is less than or equal to $d$. Is this because we can think of $x^d=1$ as being a polynomial in $F[x]$: $x^d-1=0\in F[x]$ and thus the number of distinct roots must be less than or equal to the degree? – Grifball Jul 09 '23 at 14:56
  • @Grifball Yes, that's exactly why. – zaq Jul 14 '23 at 08:50
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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 $p$.

mercio
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    That argument depends on the structure of abelian groups. – lhf Jun 17 '14 at 18:17
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    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
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    @MickG : as lhf pointed out, I rely on the structure theorem of finite commutative groups. For such a group $G$ there is a sequence of integers $(d_i)$ where $d_i$ divides $d_{i+1}$ such that $G = \oplus \Bbb Z/d_i \Bbb Z$. In the case of noncyclic groups this sequence has more than $1$ element so it has a subgroup of the form $(\Bbb Z/d_1 \Bbb Z)^2$ – mercio Jun 17 '14 at 18:41
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    @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
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    @Pavel Coupek, I am not understanding your argument for existence of an element of order |G|. – Martund Aug 23 '19 at 03:24
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    @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 03:55
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    @Martund Now, if $G$ did not have an element of order $n$, it would have at most $\sum_{d \mid n, d \neq n}\varphi(d) <\sum_{d \mid n}\varphi(d)=n$ elements, contradiction. – Pavel Čoupek Aug 23 '19 at 03:57
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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.

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