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I'm having some trouble with the following question:

Let $R$ be a ring with order $p$, where $p$ is prime. Prove that $R$ is comutative.

Because $(R,+)$ is a group then because of Lagrange's Theorem, $\exists k \in R$ such that: $$R=\{0,k,2k,...,(p-1)k\}$$

So any element of $R$ can be written as: $mk$ for $m\in \{0,...,p-1\}$, but I'm having some trouble continuing the proof from here. How can I prove this?

Eduardo Magalhães
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    This question does not need the "alternative proof" tag, since you're seeking for a proof, not a different proof. – PrincessEev Jan 19 '22 at 19:47
  • @EeveeTrainer That's why I'm asking for an "Alternative proof". See the previous comment. There is already one, and I want an alternative one. – Eduardo Magalhães Jan 19 '22 at 19:49
  • @EduardoMagalhães There does not seem to be a previous comment to see. You seem to be asking a question about the other post's suggested solution now. It is unfortunate that a) the question is old and contextless, b) the most obvious solution is only alluded to in the comments and c) the existing solution does not explain the most obvious solution. – rschwieb Jan 19 '22 at 20:14
  • Actually there are numerous questions with solutions that do explain: i've linked an additional one. – rschwieb Jan 19 '22 at 20:17

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You're almost there. Now take $p,q\in R$. $p=mk, q=nk, n,m\in\Bbb{Z}$. So: $pq=mknk=nkmk=qp$, where second equality is because elements of $\Bbb{Z}$ commute with every element in $R$ and $k$ commutes with itself.

Math101
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  • Why is $mknk=nkmk$? $m$ is not an element of the ring, so it can't commute. $mk$ is not $m\cdot k$, it's just a notation for $\underbrace{k+k+...+k}_m$. Is $km$ Even defined? – Eduardo Magalhães Jan 19 '22 at 19:50
  • @EduardoMagalhães As written, elements in $\Bbb{Z}$ commute with every element in R. $mk=k+...+k=km$. So: $nkmk=nmkk=mnkk=mknk$ – Math101 Jan 19 '22 at 19:51