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

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    $\begingroup$ This might help. $\endgroup$
    – Burak
    Jun 17, 2014 at 17:40
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    $\begingroup$ 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. $\endgroup$
    – Lubin
    Jun 17, 2014 at 17:48
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    $\begingroup$ 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 :) $\endgroup$
    – rschwieb
    Jun 17, 2014 at 17:54
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    $\begingroup$ Keith Conrad has collected six proofs of this theorem for the field $\mathbb Z/p$ at math.uconn.edu/~kconrad/blurbs/grouptheory/cyclicFp.pdf. $\endgroup$
    – lhf
    Jun 17, 2014 at 18:40
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    $\begingroup$ 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. $\endgroup$
    – MickG
    Jun 18, 2014 at 12:37

3 Answers 3

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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.

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    $\begingroup$ 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? $\endgroup$
    – Grifball
    Jul 9, 2023 at 14:56
  • $\begingroup$ @Grifball Yes, that's exactly why. $\endgroup$
    – zaq
    Jul 14, 2023 at 8:50
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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 $p$.

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    $\begingroup$ That argument depends on the structure of abelian groups. $\endgroup$
    – lhf
    Jun 17, 2014 at 18:17
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    $\begingroup$ Uh-huh. And how do you prove that «the only commutative finite groups with this property are the cyclic groups»? $\endgroup$
    – MickG
    Jun 17, 2014 at 18:33
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    $\begingroup$ @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 $(\Bbb Z/d_1 \Bbb Z)^2$ $\endgroup$
    – mercio
    Jun 17, 2014 at 18:41
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    $\begingroup$ @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 $|G|$. $\endgroup$ Jun 17, 2014 at 20:02
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    $\begingroup$ @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. $\endgroup$ Aug 23, 2019 at 3:57
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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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