# 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.

$\blacksquare$