Proving $\phi(G)$ is cyclic if $\phi$ : $G \to H$ is group homomorphism and $G$ is cyclic [duplicate]

Qusetion: Prove $\phi(G)$ is cyclic if $\phi$ : $G \to H$ is group homomorphism and $G$ is cyclic

Do I do something like this to prove this:
Let $G$ be a cyclic group and $G$=$\langle g \rangle$
Then $\phi$ : $\mathbb Z \to \langle g \rangle$ by $\phi(n)$= $g^{n}$, $n\in \mathbb Z$
I am not sure of my approach to this. Need some help

Solution 1:

Since $G$ is cyclic, it is generated by some element $g \in G$. Let us take a closer look at $\phi(G)$: suppose $h \in \phi(G)$. This means that there is some $x \in G$ such that $\phi(x) = h$. Since $g$ generates $G$, we have that $x = g^k$ for some $k \in \mathbb{Z}$. Because of the definition of homomorphism, we have that $$h = \phi(x) = \phi(g^k) = \phi(g)^k$$ so $\phi(G)$ is generated by $\phi(g)$, showing that $\phi(G)$ is cyclic.

Solution 2:

Hint: $\phi$ acts on G. We have $\phi(g^n)=\phi(g)^n$.

Solution 3:

Hint: See that if $G$ is generated by $g$ then its homomorphic image $\phi(G)$ is generated by $\phi(g)$.

Use property of homomorphism to prove the above fact.

Take an element $b\in \phi(G)$ then $\phi(a) =b,a=g^n$ then $b=\phi(g^n)=\phi(g)^n$.

so $\phi(G)$ is generated by $\phi(g)$, showing that $\phi(G)$ is cyclic.