Kernel of Homomorphism on Cyclic Group

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

Let $H$ be a group.

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

Let $\map \ker \phi$ denote the kernel of $\phi$.

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

Then:
 * $\map \ker \phi = \gen {g^m}$

where:
 * $m = 0$ if $\Img \phi$ is an infinite cyclic group
 * $m = \order {\Img \phi}$ if $\Img \phi$ is a finite cyclic group.

Proof
From Kernel of Group Homomorphism is Subgroup and Subgroup of Cyclic Group is Cyclic:
 * $\exists m \in \N: \map \ker \phi = \gen {g^m}$

From Homomorphic Image of Cyclic Group is Cyclic Group:
 * $\Img \phi$ is a cyclic group generated by $\map \phi g$.

Case $1$: $\Img \phi$ is infinite
$m \ne 0$.

Then $g^m$ is not the identity.

Thus:

However this contradicts $g^m \in \gen {g^m} = \map \ker \phi$.

Hence we must have $m = 0$.

Case $2$: $\Img \phi$ is finite
From Order of Cyclic Group equals Order of Generator, the order of $\map \phi g$ is $\order {\Img \phi}$.

Let $n \in \Z$.

By Division Theorem:
 * $\exists q, r \in \Z: 0 \le r < \order {\Img \phi}: n = q \order {\Img \phi} + r$

We have:

From definition of order of group element:
 * $0 < r < \order {\Img \phi} \implies \paren {\map \phi g}^r \ne e_H$

Hence:

and we see that $m = \order {\Img \phi}$ satisfies the above relation.