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If we consider $(\mathbb{Z}/120\mathbb{Z})^{\times}$ ?

Using the CRT we have : $(\mathbb{Z}/120\mathbb{Z})^{\times} \simeq (\mathbb{Z}/2^3\mathbb{Z})^{\times} \ \times (\mathbb{Z}/3\mathbb{Z})^{\times} \ \times (\mathbb{Z}/5\mathbb{Z})^{\times}$

By other properties $(\mathbb{Z}/120\mathbb{Z})^{\times}\simeq \mathbb{Z}/2\mathbb{Z}\ \times\mathbb{Z}/2\mathbb{Z}\ \times \mathbb{Z}/2\mathbb{Z}\ \times \mathbb{Z}/4\mathbb{Z}$

But we know that $(\mathbb{Z}/2^3\mathbb{Z})^{\times}$ is not cyclic and moreover $\gcd(2,2,2,4)\neq 1$ so the product cannot be a cyclic group. Does that mean that $(\mathbb{Z}/120\mathbb{Z})^{\times}$ is not cyclic too ?

Thanks in advance !

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  • $\begingroup$ For $n=120$ it is not cyclic, as you have shown. $\endgroup$ – Dietrich Burde Mar 6 '17 at 20:30
  • $\begingroup$ Short answer: what you did is correct! For a more general result: see the comment above ;) Moreover, the last step is not needed (with the order) since you have found a non cyclic subgroup of a group (and therefore the group itself is not cyclic) $\endgroup$ – Student Mar 6 '17 at 20:31
  • $\begingroup$ @Student thank you in fact it's a trap because if the group is not cyclic you cannot apply a powerful result on the number of solutions of $x^k \equiv 1 \pmod{ n}$. You will have to use CRT and it will be longer. $\endgroup$ – Maman Mar 6 '17 at 20:36
  • $\begingroup$ I do not really get your comment about the trap... Is it about my comment about the redundancy of your last line (about the order)? $\endgroup$ – Student Mar 6 '17 at 20:38
  • $\begingroup$ @Student There is a powerful lemma which says if $(Z/nZ)^{\times}$ is cyclic then the number of solutions of $x^k \equiv 1 \pmod n$ is $\gcd(k,\phi(n))$ $\endgroup$ – Maman Mar 6 '17 at 20:45