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
Aiming for a contradiction, suppose $m \ne 0$.
Then $g^m$ is not the identity.
Thus:
\(\ds \map \phi {g^m}\) | \(=\) | \(\ds \paren {\map \phi g}^m\) | Homomorphism of Power of Group Element | |||||||||||
\(\ds \) | \(\ne\) | \(\ds e_H\) | Definition of Infinite Cyclic Group |
However this contradicts $g^m \in \gen {g^m} = \map \ker \phi$.
Hence we must have $m = 0$.
$\Box$
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:
\(\ds \map \phi {g^n}\) | \(=\) | \(\ds \paren {\map \phi g}^n\) | Homomorphism of Power of Group Element | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \phi g}^{q \order {\Img \phi} + r}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {\map \phi g}^{\order {\Img \phi} } }^q \paren {\map \phi g}^r\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds {e_H}^q \paren {\map \phi g}^r\) | as $\paren {\map \phi g}^{\order {\Img \phi} } = e_H$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map \phi g}^r\) | Power of Identity is Identity |
From definition of order of group element:
- $0 < r < \order {\Img \phi} \implies \paren {\map \phi g}^r \ne e_H$
Hence:
\(\ds \map \phi {g^n} = e_H\) | \(\iff\) | \(\ds r = 0\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds g^n \in \ker \phi\) | \(\iff\) | \(\ds n = q \order {\Img \phi}\) | Definition of Kernel of Group Homomorphism | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds g^n \in \gen {g^m}\) | \(\iff\) | \(\ds g^n \in \gen {g^\order {\Img \phi} }\) | Group Generated by Singleton | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \gen {g^m}\) | \(=\) | \(\ds \gen {g^\order {\Img \phi} }\) | Definition of Set Equality |
and we see that $m = \order {\Img \phi}$ satisfies the above relation.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $7$: Homomorphisms: Exercise $8$