Is this particular group cyclic? [duplicate]

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 ?

• For $n=120$ it is not cyclic, as you have shown. – Dietrich Burde Mar 6 '17 at 20:30
• @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. – Maman Mar 6 '17 at 20:36
• @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 $\gcd(k,\phi(n))$ – Maman Mar 6 '17 at 20:45