Show that $\mathbb{Z}/5\mathbb{Z} = \langle a\rangle$

Question

Show that $\mathbb{Z}/5\mathbb{Z} = \langle a\rangle$.

I have that $a$ is an element of the non zero complex numbers $\mathbb{C}^*$ and $a^{5}=1$ where $a \neq 1$. I have shown already that $|a| = 5$ and the function that maps $(\mathbb{Z},+) \to (\mathbb{C}^*,\times)$ given by $n \mapsto a^n$ is a homomorphism.

How do I now go on to show that $\mathbb{Z}/5\mathbb{Z} = \langle a\rangle$?

Answer 1

Denote your homomorphism by $\varphi$. Using the fact that the order of $a$ is $5$, one can show that Ker $\varphi$ = $5\mathbb{Z}$. Now, the fundamental homomorphism theorem for groups says that $\mathbb{Z}/$Ker $\varphi$ $\cong$ Im $\varphi$. Now we have Im $\varphi = ⟨a⟩$ such that $\mathbb{Z}$/ $5\mathbb{Z} \cong ⟨a⟩$.

Answer 2

Actually $\mathbb{Z}/5\mathbb{Z}$ is a cyclic group of order $5$, and since $5$ is prime then all the non-identity elements of $\mathbb{Z}/5\mathbb{Z}$ are of order $5$ then all the non-identity elements are generators of this group. Now, $a$ is an element of $\mathbb{C}^*$ and a is not identity and $a^5=1$ then $|a|=5$ and $<a>$ is a cyclic group of order $5$ and having the elements $\{a,a^2,a^3,a^4,1\}$ where all the elements are of order $5$. So, $\{a,a^2,a^3,a^4,1\}$ is isomorphic to $\mathbb{Z}/5\mathbb{Z}$, Hence , $\mathbb{Z}/5\mathbb{Z}=<a>$. Is my way correct or convincing?