Translation Mapping is Bijection

Theorem

Let $\struct {G, +}$ be an abelian group.

Let $g \in G$.

Let $\tau_g: G \to G$ be the translation by $g$:

$\forall h \in G: \map {\tau_g} h = h + \paren {-g}$

where $-g$ is the inverse of $g$ with respect to $+$ in $G$.

Then $\tau_g$ is a bijection.

$\ds \forall h_1, h_2 \in G: \, $

$\ds \map {\tau_g} {h_1}$

$=$

$\ds \map {\tau_g} {h_2}$

$\ds \leadsto \ \ $

$\ds h_1 + \paren {-g}$

$=$

$\ds h_2 + \paren {-g}$

Definition of $\tau_g$

$\ds \leadsto \ \ $

$\ds h_1$

$=$

$\ds h_2$

$\Box$

$\ds \forall h_1 \in G: \exists h_2 \in G: \, $

$\ds h_1 + g$

$=$

$\ds h_2$

$\ds \leadsto \ \ $

$\ds h_1$

$=$

$\ds h_2 + \paren {-g}$

$\ds \leadsto \ \ $

$\ds h_1$

$=$

$\ds \map {h_2} g$

Definition of $\tau_g$

$\blacksquare$