Homomorphic Image of Group Element is Coset
Theorem
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Let $h \in H$.
Then $\Preimg h$ is either the empty set or a coset of $\map \ker \phi$.
Proof
There are two possibilities for any $h \in H$.
- $(1): \quad \Preimg h = \O$
- $(2): \quad \Preimg h \ne \O$
If $(1)$, then the first disjunct of the result is satisfied.
Now suppose $(2)$ holds.
Let $e_G$ and $e_H$ be the identity elements of $G$ and $H$ respectively.
Let $K = \map \ker \phi$.
Let $x, y \in G$ such that $\map \phi x = \map \phi y$.
Then:
\(\ds e_H\) | \(=\) | \(\ds \map \phi x * \paren {\map \phi y}^{-1}\) | Definition of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x * \map \phi {y^{-1} }\) | Homomorphism to Group Preserves Inverses | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {x \circ y^{-1} }\) | $\phi$ is a homomorphism | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x \circ y^{-1}\) | \(\in\) | \(\ds K\) | Definition of Kernel of Group Homomorphism | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds y K\) | Element in Left Coset iff Product with Inverse in Subgroup |
Thus:
- $\set {x: \map \phi x = \map \phi y}$ is a subset of $y K$.
From Kernel is Normal Subgroup of Domain we have that $y K = K y$.
Now suppose $x \in K y$.
Then, by definition, $x = k y$ for some $k \in K$.
Thus:
\(\ds x\) | \(=\) | \(\ds k y\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \phi x\) | \(=\) | \(\ds \map \phi k * \map \phi y\) | $\phi$ is a homomorphism | ||||||||||
\(\ds \) | \(=\) | \(\ds e_H * \map \phi y\) | Definition of Kernel of Group Homomorphism: $k \in K$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi y\) | Definition of Identity Element |
A similar process gives that $x \in y K \implies \map \phi x = \map \phi y$.
Hence the result.
$\blacksquare$
Sources
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.10$: Theorem $22 \ (2)$