Cancellable Element is Cancellable in Subset
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Theorem
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \circ} \subseteq \struct {S, \circ}$.
Let $x \in T$ be cancellable in $S$.
Then $x$ is also cancellable in $T$.
Proof
Let $x \in T$ be cancellable in $S$.
Then by definition, $x \in T$ is left cancellable in $S$.
It follows from Left Cancellable Element is Left Cancellable in Subset that $x \in T$ is left cancellable in $T$.
Again by definition, $x \in T$ is right cancellable in $S$.
It follows from Right Cancellable Element is Right Cancellable in Subset that $x \in T$ is right cancellable in $T$.
Thus $x$ is also both left cancellable and right cancellable in $T$.
So by definition, $x$ is cancellable in $T$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets