# Constant Mapping to Identity is Homomorphism/Groups

< Constant Mapping to Identity is Homomorphism(Redirected from Constant Mapping to Group Identity is Homomorphism)

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## Theorem

Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identities $e_1$ and $e_2$ respectively.

Let $\phi_e: \struct {G_1, \circ_1} \to \struct {G_2, \circ_2}$ be the constant mapping defined as:

- $\forall x \in G_1: \map {\phi_e} x = e_2$

Then $\phi_e$ is a group homomorphism whose image is $\set {e_2}$ and whose kernel is $G_1$.

## Proof

Let $x, y \in G_1$.

Then:

\(\ds \map {\phi_e} {x \circ_1 y}\) | \(=\) | \(\ds e_2\) | as $x \circ_1 y \in G_1$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \map {\phi_e} x \circ_2 \map {\phi_e} y\) | as $\map {\phi_e} x = e_2$ and $\map {\phi_e} y = e_2$ |

So $\phi_e$ is a group homomorphism.

The results about image and kernel follow directly by definition.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): Chapter $7$: Homomorphisms: Exercise $1$