Find the structure of $\mathbb{Z}_{120}^*$

Question

How to find the structure (in term of cyclic groups) of $\mathbb{Z}_{120}^*$?

I know that the number of elements of $\mathbb{Z}_{120}^*= \phi(120) = 32 = 2^5$

But then, any hints?

Answer 1

I want to thank Tobias Kildetoft for the useful suggestion of using the Chinese Remainder Theorem.

So we have that $120 = 8 \; 3 \; 5$

$\mathbb{Z}_{120} = \mathbb{Z}_{3} \times \mathbb{Z}_{8} \times \mathbb{Z}_{5}$

And from this

$\mathbb{Z}_{120}^* = \mathbb{Z}_{3}^* \times \mathbb{Z}_{8}^* \times \mathbb{Z}_{5}^*$

Now, from a general result: $\mathbb{Z}_{p}^* = \mathbb{Z}_{p-1}$ if $p$ is a prime integer.

$\mathbb{Z}_{8}^*$ has $4$ elements, it could be either $\mathbb{Z}_{4}$ or $\mathbb{Z}_{2} \times \mathbb{Z}_{2}$. But it is not cyclic since $\mathbb{Z}_{8}^* = \{1,3,5,7 \}$ and so we see it has no element of order $4$ , and so we have the desired decomposition:

$\mathbb{Z}_{120}^* = \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{4}$