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How to prove Euler's theorem using Lagrange's theorem?

If $a$ and $n$ are relatively prime then $a^{\phi(n)}\equiv 1 \pmod n$

Wikipedia says that it can done and that $\phi(n)$ is the order of the multiplicative group of integers modulo $n$. But I'm not sure how to proceed with the proof. Any ideas/hints?

  • This is a duplicate, so as much as I would want to provide my input, it has already been done justice. Lagrange's theorem is simply overkill by the way. –  May 24 '18 at 22:48
  • @JefferyOpoku-Mensah I am aware of the non-group theory proof, but wanted to know the one which uses group theory –  May 24 '18 at 22:50
  • Ah. I should say the proof is still group-theoretic. It is pretty much the restriction of Lagrange's theorem to abelian groups in fact, so the details carry over, except the argument is clouded with the one line phrase "Lagrange's theorem." –  May 24 '18 at 22:52

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Since $\gcd(a,n)=1$ we have that $\overline{a}\in(\mathbf{Z}/n\mathbf{Z})^*$. The order of $\overline{a}$ must divide the order of the group, which is $\phi(n)$. This gives $\overline{a}^{\phi(n)}=\overline{1}$, or in mod notation $a^{\phi(n)}=1\text{ mod }n$.

rae306
  • 9,850
  • Are you referring to the multiplicative group by $(\mathbf{Z}/n\mathbf{Z})^*$ ? –  May 24 '18 at 22:51
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    Yes, exactly :) – rae306 May 24 '18 at 22:52
  • Note that the argument 'the order of $\overline{a}$ must divide the order of the group' is simply Lagrange applied on the subgroup $\langle\overline{a}\rangle$. – rae306 May 24 '18 at 22:58