Homomorphic Image of Group Element is Coset

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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 $\map \ker \phi$.


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 $\map \phi x = \map \phi y$.


\(\ds e_H\) \(=\) \(\ds \map \phi x * \paren {\map \phi y}^{-1}\) 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


$\set {x: \map \phi x = \map \phi y}$ 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 $k \in K$.


\(\ds x\) \(=\) \(\ds k y\)
\(\ds \leadsto \ \ \) \(\ds \map \phi x\) \(=\) \(\ds \map \phi k * \map \phi y\) $\phi$ 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.