Let $m$ be a positive integer, let $\chi$ be the Dirichlet character on $\mathbb{Z}/m\mathbb{Z}$ which means $\chi$ is a group homomorphism from $(\mathbb{Z}/m\mathbb{Z})^{*}$ to $\mathbb{C}^{*}$. We know all character form a group denoted as $X_{m}$ and it is isomorphism to $(\mathbb{Z}/m\mathbb{Z})^{*}$. Let $p$ be a prime number which is coprime with $m$. Then we have group isomorphism $$X_{m}\to \mathbb{C}^{*}: \chi\to \chi (p)$$
The image form a group.
My question is why the image is a cyclic group?
I think this question should be easy but now I have no idea. Could you help me?
By Euler's theorem, $p^{\varphi(m)}\equiv1\pmod m$, so $\chi(p)$ belongs to
$$ G:=\left\{x\in\mathbb C\;\middle|\; x^{\varphi(m)}=1\right\}=\left\{\exp\left(\frac{2\pi i}{\varphi(m)}\right)\;\middle|\; i=0,\ldots,\varphi(m)-1\right\}. $$ Since $G$ is cyclic, the image of the homomorphism, as a subgroup of a cyclic group, is cyclic.
Hope this helps.