Isomorphism Preserves Left Cancellability

Theorem

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.

Then:

$a \in S$ is left cancellable in $\struct {S, \circ}$

if and only if:

$\map \phi a \in T$ is left cancellable in $\struct {T, *}$.

Let $a$ be left cancellable in $\struct {S, \circ}$.

Let $x \in S$ and $y \in S$ be arbitrary.

Then:

$\ds \map \phi a * \map \phi x$

$=$

$\ds \map \phi a * \map \phi y$

$\ds \leadsto \ \ $

$\ds \map \phi {a \circ x}$

$=$

$\ds \map \phi {a \circ y}$

Morphism Property

$\ds \leadsto \ \ $

$\ds \map \phi x$

$=$

$\ds \map \phi y$

as $a$ is left cancellable

That is, $\map \phi a$ is left cancellable in $\struct {T, *}$.

As $\phi$ is an isomorphism, then so is $\phi^{-1}$.

So the same proof works in reverse in exactly the same way.

$\blacksquare$

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.6$