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I'm trying to find a proof of this:

The group $\langle\mathbb{Z}_n,\oplus\rangle$ is cyclic for every $n$, where $1$ is a generator. The generators of the group $\langle\mathbb{Z}_n,\oplus\rangle$ are all $g \in \mathbb{Z}_n $ for which $\gcd(g,n)=1$, as the reader can prove as an exercise.

It is perfectly clear that $1$ generates all $\mathbb{Z}_n$, but I can't get myself to understand the second part or find a way to prove it. Thanks.


marked as duplicate by Dietrich Burde, Zev Chonoles abstract-algebra Jun 19 '15 at 13:08

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  • $\begingroup$ It is proved in the answer of user134824 here. $\endgroup$ – Dietrich Burde Jun 19 '15 at 13:02
  • $\begingroup$ thanks a lot for the referral! now also the role of the bezout identity in the proof is clear! $\endgroup$ – f1sh3r0 Jun 19 '15 at 13:13


Use Bézout's identity to prove that if $\,\gcd(g,n)=1$, $1$ can be obtained as a multiple of $g$ (in $\mathbf Z_n$).

  • $\begingroup$ I had a quick look at the bezout identity, but isn't there a simpler way? the text i quoted was from a first year undergrad discrete mathematics script, and the identity was never mentioned before in the text so i doubt this is the way the author imagined the proof, thank you though! $\endgroup$ – f1sh3r0 Jun 19 '15 at 12:56
  • $\begingroup$ Isn't Bézout's identity seen in high school? I'll think of a proof without it, but I won't be necessarily as simple… $\endgroup$ – Bernard Jun 19 '15 at 13:31
  • $\begingroup$ i didn't see it, but the referral to another question helped me understand the whole proof including Bézout's identity, so the question is closed - thanks again! $\endgroup$ – f1sh3r0 Jun 19 '15 at 13:35
  • $\begingroup$ Anyway, a direct proof would probably consist in proving a particular version of Bézout, i.e. reinventing the wheel. $\endgroup$ – Bernard Jun 19 '15 at 13:37

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