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Edit. By "group-theoretic" I don't mean "avoiding one of the two operations of the field" (the very claim statement uses both!), but rather avoiding any field-specific property/lemma to prove the claim, like the linked proof, on the other hand, does.

The following claim has a one-liner proof, relying on the fact that $\mathbb Z/p\mathbb Z$ ($p$ odd prime) is a field. My question is to whether the proof below really entirely lies within group theory, as it seems to me, or whether field theory enters here too, at some point that I'm overlooking.

Claim. For every $\alpha\in\mathbb Z/p\mathbb Z$, $\alpha\ne0,1$: $$1+\alpha+\dots+\alpha^{m_\alpha-1}=0$$ where $m_\alpha$ is the multiplicative order of $\alpha$.

Proof based on $\mathbb Z/p\mathbb Z$ being a field: see here. $\Box$

Group-theoretic(?) proof:

For $p$ an odd prime, any automorphism $\varphi$ of the additive group $\mathbb Z/p\mathbb Z$ fixes $0$ and permutes the elements of $X:=\mathbb Z/p\mathbb Z\setminus\{0\}$ (as automorphisms preserve elements' order). If $\varphi\ne id$, necessarily $\varphi(1)\ne 1$, as otherwise $\varphi(k)=k\varphi(1)=k$ for every $k\in\mathbb Z/p\mathbb Z$, hence $\varphi= id$: contradiction; therefore, for every $k\in X$, $\varphi(k)\ne k$, as otherwise from $\varphi(k)= k$ for some $k\in X$ would follow $\varphi(1)=1$, a contradiction. So, $\sigma:=\varphi_{|X}$ moves all the elements of $X$. If $m$ is the multiplicative order of $\sigma(1)$, then $\sigma$ has the $m$-cycle $(1,\sigma(1),\dots,\sigma(1)^{m-1})$ among its disjoint factors. Let's suppose $k:=1+\sigma(1)+\dots+\sigma(1)^{m-1}\ne 0$; then: \begin{alignat}{1} \varphi(k) &= \sigma(1+\sigma(1)+\dots+\sigma(1)^{m-1}) \\ &= \sigma(1)+\sigma(1)^2+\dots+\sigma(1)^{m} \\ &= 1+\sigma(1)+\sigma(1)^2+\dots+\sigma(1)^{m-1} \\ &= k \end{alignat} a contradiction. Therefore, $1+\sigma(1)+\sigma(1)^2+\dots+\sigma(1)^{m-1}=0$. $\Box$

Kan't
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    I can't see immediately why the cycle is $(1,\sigma(1),\sigma(1)^2, \dots, \sigma(1)^{m-1})$; I thought it was $(1,\sigma(1),\sigma^2(1), \dots, \sigma^{m-1}(1))$. And later why $\sigma(\sigma(1))=\sigma(1)^2$. – ancient mathematician Feb 07 '25 at 16:15
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    This theorem doesn't make sense for a group because we require both multiplication and addition. – John Douma Feb 07 '25 at 16:23
  • Strictly speaking, you are right, @JohnDouma. I meant "modulo that fact". I just noticed that in the "group" version we don't use fields' fact: $xy=y\iff (y\ne 0\wedge x=1)\vee(y=0)$. – Kan't Feb 07 '25 at 17:17
  • The group theoretic part is just $\alpha^{m_\alpha}=1$ by definition of $m_\alpha$. Then $0=\alpha^{m_\alpha}-1=(\alpha-1)(1+\alpha+\dots+\alpha^{m_\alpha-1})$ but $\alpha-1\ne0$ by hypothesis. The last part needs that $\alpha$ lies in a domain. – lhf Feb 07 '25 at 17:17
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    @ancientmathematician for your latter objection, $\sigma(\sigma(1)) = \sigma(\sigma(1)\cdot 1) = \sigma(\underbrace{1+1+...+1}{\sigma(1)}) = \underbrace{\sigma(1)+...+\sigma(1)}{\sigma(1)} = \sigma(1)\sigma(1) = \sigma(1)^2$. – IAAW Feb 07 '25 at 18:33
  • @lhf $,\alpha$ needn't be in a domain. Rather, we only need that $\alpha!-!1$ is a cancellable (regular) element of a ring - see this 15 year old answer which proves a more general equality and equivalence. – Bill Dubuque Feb 07 '25 at 20:41
  • $,(\alpha!-!1)S=0!\iff! \alpha S = S.,$ Said simply: the first method cancels $,\alpha !-! 1,$ from the first equation. The second method cancels $,S,$ from the second (in group theory language). $\ \ $ – Bill Dubuque Feb 07 '25 at 21:08
  • Cancelling $S$ leaves $\alpha=1$, which is a contradiction, isn't it, @BillDubuque? – Kan't Feb 07 '25 at 21:28
  • @Kan't Yes that is what I meant, i..e. if $,S\neq 0,$ then cancelling it from $,\alpha S = S,$ yields $,\alpha =1,,$ contra hypothesis. Thus $,S = 0.\ \ $ – Bill Dubuque Feb 07 '25 at 23:32
  • Okay, but that's exactly the field argument in the linked proof that I'm trying to avoid, @BillDubuque. – Kan't Feb 08 '25 at 02:25
  • If "prime" is defined the way algebraists define it, then what you are trying to prove just follows from that definition and $(1-\alpha)(1+\cdots+\alpha^{m-1})=1-\alpha^m$. So can you clarify your definition of "prime" for $p$ here? – 2'5 9'2 Feb 08 '25 at 03:30
  • As long as you are calculating the sum of elements in a set stable under multiplication by some element you can make some deductions. You do need the field axioms to conclude that the sum must be zero. For example in the ring $\Bbb{Z}_8$ the subsets $M={1,3}$ and $N={2,6}$ are both stable under multiplication by $3$. If $S(M)$ and $S(N)$ are the sums, it follows that $(3-1)S(M)=0=(3-1)S(N)$. Here $3-1$ is a zero divisor, and you see what the problem is. – Jyrki Lahtonen Feb 08 '25 at 05:33
  • My point is that you don't need to consider powers of an element. Any set stable under multiplication will also work. And also that stability under multiplication by $\alpha$ won't suffice, if $\alpha-1$ is a zero divisor. – Jyrki Lahtonen Feb 08 '25 at 05:39
  • By "prime" I mean a prime number, @2'59'2. – Kan't Feb 08 '25 at 05:54
  • @Kan't I'm serious. Define "prime number" as you understand it. If you truly mean "prime", this result is trivial. I suspect you will offer a definition for "irreducible", and then I was going to ask if you accept citing the fundamental theorem of arithmetic which makes irreducible equivalent to prime. – 2'5 9'2 Feb 08 '25 at 06:08
  • Sorry, I didn't mean to be ironic at all, @2'59'2, and yes, the result is trivial as the linked answer shows. But that is not the point of my question. – Kan't Feb 08 '25 at 06:16
  • Ok, I'll leave at this: it is trivial without citing field facts, which it sounds like was your original objective. Only posting comments since I know this doesn't answer your question. – 2'5 9'2 Feb 08 '25 at 06:57

1 Answers1

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Your proof seems awkward to me, as written it's very unclear what it's really using. It uses two facts both of which are extremely specific to the group $\mathbb{Z}/p\mathbb{Z}$:

  1. The automorphism group of $\mathbb{Z}/p\mathbb{Z}$ acts transitively on nonzero elements.
  2. Every non-identity automorphism has no nonzero fixed points.

These both follow from the fact that the automorphisms are given by multiplication by an element of $(\mathbb{Z}/p\mathbb{Z})^{\times}$. Both of them are required to make your argument work; you need fact 1 in the last step and fact 2 to get the contradiction.

And you do not prove either of them! I think you'll find that in order to do so you need to prove something more or less equivalent to $\mathbb{Z}/p\mathbb{Z}$ being a field.

It's also not hard to see that the claim is false over rings which are not integral domains. The universal counterexample is the group ring

$$R = \mathbb{Z}[C_m] \cong \mathbb{Z}[g]/(g^m - 1)$$

where by construction $g$ is an element of multiplicative order $m$ but

$$1 + g + \dots + g^{m-1} \neq 0.$$

In fact $R$ admits a homomorphism $R \to \mathbb{Z}$ sending $g \mapsto 1$, so the image of $1 + g + \dots + g^{m-1}$ under this homomorphism is $m \in \mathbb{Z}$.

Qiaochu Yuan
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  • I think I proved what suffices to the claim, which -yes- is specific to the group $\mathbb Z/p\mathbb Z$, which on the other hand is the subject of the claim. So I don't get what your example is a counterexample of. – Kan't Feb 07 '25 at 22:25
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    @Kan't: you did not prove either of the facts listed above, and you need both of them to complete your argument. I am claiming that in order to do so you need to prove something that is equivalent to $\mathbb{Z}/p\mathbb{Z}$ being a field, and separately, as evidence that this is true, I gave a counterexample over a non-field. – Qiaochu Yuan Feb 07 '25 at 22:28
  • Okay. I'm struggling to see: as for 1, why we need the transitivity (which is true) to show that $\sigma$ has the cycle made of the powers of $\sigma(1)$ (which suffices to my argument), and as for 2, what's wrong with the part where I precisely show that $\sigma(k)\ne k$ for every $k\in X$ (starting from "If $\varphi \ne id$" onwards). – Kan't Feb 08 '25 at 02:14
  • I've edited this second point, though. – Kan't Feb 08 '25 at 02:35
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    @Kan't Your argument $\phi(k)=k\implies \phi(1)=1$ basically relies on the fact that $1$ is a multiple of $k$, which is equivalent to $k$ having an inverse (as we are in the prime field), so you are proving that $\Bbb{Z}/p\Bbb{Z}$ is a field. – Jyrki Lahtonen Feb 08 '25 at 06:23
  • Thank you, @JyrkiLahtonen. *That* is the point I was overlooking – Kan't Feb 08 '25 at 07:11
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    @Kan't: similarly your implicit claim at the very end that $\sigma(1)$ can take on every possible nonzero value $k$. It's clear that an endomorphism of $\mathbb{Z}/p\mathbb{Z}$ can do this, namely multiplication by $k$. But to prove that multiplication by $k$ is an automorphism you have to prove that $k$ has a multiplicative inverse, again. So both of the facts above require this. – Qiaochu Yuan Feb 08 '25 at 17:37