Homomorphic Image of Cyclic Group is Cyclic Group

Theorem
Let $G$ be a cyclic group with generator $g$.

Let $H$ be a group.

Let $\phi: G \to H$ be a (group) homomorphism.

Let $\Img G$ denote the homomorphic image of $G$ under $\phi$.

Then $\Img G$ is a cyclic group with generator $\map \phi g$.

That is:
 * $\phi \sqbrk {\gen g} = \gen {\map \phi g}$

Proof
Let $y \in \Img G$.

Then $\exists x \in G: y = \map \phi x$.

As $G$ be a cyclic group with generator $g$, $x = g^n$ for some $n \in \Z$.

Thus by Homomorphism of Power of Group Element:
 * $y = \paren {\map \phi g}^n$

and so is a power of $\map \phi g$.

As $y$ is arbitrary, it follows that all elements of $\Img G$ are powers of $\map \phi g$.

Hence the result, by definition of cyclic group.