If $o(\mathrm{a})$ be finite, $\langle a\rangle\simeq \Bbb {Z}/m\Bbb{Z}$ for some positive integer $m$ using Isomorphism theorem.

Question

Let $G$ be a group. A mapping $\phi: \mathbb{Z} \rightarrow G$ is defined by $\phi(n)=a^{n},$ where $a \in G$ and $n \in \mathbb{Z}$ show that $\phi$ is a homomorphism. Moreover, if $o(a)$ be finite, $\langle a\rangle\simeq \frac{\mathbb{Z}}{m\mathbb{Z}}$ for some positive integer $m$ using Isomorphism theorem.

My attempt:

First part : For $m,n\in \mathbb{Z}, \phi({m+n})=a^{m+n}=a^ma^n=\phi(m)\phi(n)\implies \phi$ is homomorphism.

How can I show the $2^{\text{nd}}$ part?

Answer 1

Let $G,H$ be groups and $\phi:G\rightarrow H$ be homomorphism. Then the First Isomorphism Theorem states that $$G/\ker \phi \cong \phi(G)$$

Let $m=o(a)$. So first you had shown that $\phi$ is a homomorphism.
Next you want to show that $\phi(\Bbb{Z})=\langle a\rangle$.
Note that $\phi(\Bbb{Z})=\{a^z:z\in \Bbb{Z}\}=\langle a\rangle$ by the definition of subgroup generated by a single element.

Lastly you have to show that $\ker \phi =m\Bbb{Z}$.
Notice that $\ker \phi=\{z\in \Bbb{Z}:a^z=1\}$.
Recall a property that $a^z=1$ if and only if $m$ divides $z$.
Hence we get $\ker\phi=m\Bbb{Z}$ and the result follows.