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In this answer to the question: Show that every group of prime order is cyclic, I couldn't understand two things:

  1. What does $a\perp b$ mean? Does it mean $a\neq b$?

  2. Why is $x^{p}=1$?

It would be very helpful if someone could clear this for me.

  • 2
    In that context $\perp$ means coprime. – Michal Adamaszek May 19 '18 at 18:28
  • And if $x^p\ne1$, what other value could $x^p$ take, given what has already been proved in this answer? Hint: since the order of the group is $p$, $x^p=x^i$ for some $i<n$. – Jean-Claude Arbaut May 19 '18 at 18:30
  • @MichalAdamaszek Thanks! Could you also answer the second part? –  May 19 '18 at 18:31
  • BTW something is missing in the assumptions of that lemma. I think it meant to be $x^a=x^b=1$. – Michal Adamaszek May 19 '18 at 18:39
  • If $p$ is the order of the group, the order of an element divides the order of the group, by Lagrange's theorem. – Bernard May 19 '18 at 18:48
  • @Jean-ClaudeArbaut I'm a bit confused because "order of group" is $p$ does not imply that the order of $x$ is $p$. Even $x^{p-3}$(say) could have been $1$ –  May 19 '18 at 18:50
  • @Blue No, you don't know that the order of $x$ is $p$ yet, but you know the group has exactly $p$ elements, and you know $1,x,x^2,\dots,x^{p-1}$ are distinct, so they are all the $p$ elements of the group. Hence $x^p=x^i$ with $i<n$ (because $x^p$ can't be another element). Which $i$? – Jean-Claude Arbaut May 19 '18 at 19:10
  • @Jean-ClaudeArbaut But how do we know $1,x,x^2,...,x^{p-1}$ are all distinct? –  May 19 '18 at 19:18
  • Because if $x^a=x^b$ with $x\ne1$ and $a<b<p$, then $x^{b-a}=1$, and that's impossible because $$ would be a strict subgroup of $G$, with more than one element. And since it's a subgroup, its order divides the order of $G$, which is prime. But a prime $p$ has no divisor except $1$ and $p$: so the hypothesis $x^a=x^b$ can't be true. So, if $x\ne1$, the elements $1,x,x^2,\dots,x^{p-1}$ are distinct. – Jean-Claude Arbaut May 19 '18 at 19:26
  • But note that here I use Lagrange's theorem (the OP of the other question explicitly required not using it). – Jean-Claude Arbaut May 19 '18 at 19:32
  • @Jean-ClaudeArbaut Thanks. I get it now. :) –  May 19 '18 at 19:58

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