Number of homomorphisms between $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{C}^{\times}$ 
Problem: Let $\mathbb{C}^{\times}$ denote the nonzero complex numbers under multiplication. Show
that there are $5$ homomorphisms from $\mathbb{Z}/5\mathbb{Z}\to \mathbb{C}^{\times}$

I tried to do the following:
Let's say $\phi :\mathbb{Z}/5\mathbb{Z} \to \mathbb{C}^{\times} $. We know that $\mid \text{ker}(\phi)\mid$ divides $\mathbb{Z}/5\mathbb{Z}$. So the kernel is either full $\mathbb{Z}/5\mathbb{Z}$ or $1_G$ only. I am not sure what to do next and how to proceed. Can someone help me out?
Any hint is highly appreciated.
Thank you!
 A: Hint: Let's write $G=\mathbb{Z}/5\mathbb{Z}$ and $g=1+5\mathbb{Z}$. Note the following:

*

*If $\varphi$ is such a homomorphism then we have $\varphi(g)^5=1$, and so $\varphi(g)$ is a fifth root of unity.


*If two such homomorphisms $\varphi,\psi$ satisfy $\varphi(g)=\psi(g)$ then $\varphi=\psi$.
The first part should help you find $5$ homomorphisms. The second part should help you prove there aren't any other homomorphisms.
A: $\newcommand{\Z}{\mathbb Z}$
$\newcommand{\C}{\mathbb C}$
I will denote by $\overline a$ the elements of $(\Z/n\Z, +)$.
You know that $\overline 1$ generates this group.
Hence, knowing the image of $\overline 1$ will leave no choice for the images of the other elements of the group !
Moreover, you know that $\overline 1$ is of order $5$, i.e. $5 \cdot \overline 1 = \overline 0$, so it is necessary that $\phi(1)^5 = 1$ in $\C$. There are exactly $5$ such numbers, namely $e^{\frac{2i\pi} 5}$ for $0 \leq i < 5$.
What you need to prove (a bit more precisely than in my semi-informal reasoning) is that this procedure correctly defines a morphism, that there are no other morphisms $\Z/n\Z \to \C^\ast$, and that they are all different.
