Subset Product within Semigroup is Associative/Corollary
Jump to navigation
Jump to search
Corollary to Subset Product within Semigroup is Associative
Let $\struct {S, \circ}$ be a semigroup.
Then:
| \(\ds x \circ \paren {y \circ S}\) | \(=\) | \(\ds \paren {x \circ y} \circ S\) | ||||||||||||
| \(\ds x \circ \paren {S \circ y}\) | \(=\) | \(\ds \paren {x \circ S} \circ y\) | ||||||||||||
| \(\ds \paren {S \circ x} \circ y\) | \(=\) | \(\ds S \circ \paren {x \circ y}\) |
Proof
From the definition of Subset Product with Singleton:
| \(\ds x \circ \paren {y \circ S}\) | \(=\) | \(\ds \set x \circ_\PP \paren {\set y \circ_\PP S}\) | ||||||||||||
| \(\ds x \circ \paren {S \circ y}\) | \(=\) | \(\ds \set x \circ_\PP \paren {S \circ_\PP \set y}\) | ||||||||||||
| \(\ds \paren {S \circ x} \circ y\) | \(=\) | \(\ds \paren {S \circ_\PP \set x} \circ_\PP \set y\) |
The result then follows directly from Subset Product within Semigroup is Associative.
$\blacksquare$
Also see
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 6.1$. The quotient sets of a subgroup