Isomorphism Preserves Left Cancellability
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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 $\struct {S, \circ}$
- $\map \phi a \in T$ is left cancellable in $\struct {T, *}$.
Proof
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.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.6$