Conditions under which Commutative Semigroup is Group/Lemma 1
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Lemma for Conditions under which Commutative Semigroup is Group
Suppose the following:
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {S, \circ}$ have the following properties:
| \((1)\) | $:$ | \(\ds \forall x \in S: \exists y \in S:\) | \(\ds y \circ x = x \) | ||||||
| \((2)\) | $:$ | \(\ds \forall x, y \in S:\) | \(\ds y \circ x = x \implies \exists z \in S: z \circ x = y \) |
Then:
- If $y \circ x = x = y' \circ x$, then $y = y'$.
Proof
| \(\ds y \circ x\) | \(=\) | \(\, \ds x \, \) | \(\, \ds = \, \) | \(\ds y' \circ x\) | by hypothesis | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {y \circ x} \circ z\) | \(=\) | \(\, \ds x \circ z \, \) | \(\, \ds = \, \) | \(\ds \paren {y' \circ x} \circ z\) | where $z \circ x = y$ from $(2)$ | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds y \circ \paren {x \circ z}\) | \(=\) | \(\, \ds x \circ z \, \) | \(\, \ds = \, \) | \(\ds y' \circ \paren {x \circ z}\) | Semigroup Axiom $\text S 1$: Associativity | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds y \circ \paren {z \circ x}\) | \(=\) | \(\, \ds z \circ x \, \) | \(\, \ds = \, \) | \(\ds y' \circ \paren {z \circ x}\) | Definition of Commutative Semigroup | ||||||||
| \(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds y \, \) | \(\, \ds = \, \) | \(\ds y' \circ y\) | from $(2)$ |
Similarly:
| \(\ds y' \circ x\) | \(=\) | \(\, \ds x \, \) | \(\, \ds = \, \) | \(\ds y \circ x\) | by hypothesis | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {y' \circ x} \circ z'\) | \(=\) | \(\, \ds x \circ z' \, \) | \(\, \ds = \, \) | \(\ds \paren {y \circ x} \circ z'\) | where $z' \circ x = y'$ from $(2)$ | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds y \circ \paren {x \circ z'}\) | \(=\) | \(\, \ds x \circ z' \, \) | \(\, \ds = \, \) | \(\ds y' \circ \paren {x \circ z'}\) | Semigroup Axiom $\text S 1$: Associativity | ||||||||
| \(\ds \leadsto \ \ \) | \(\ds y \circ \paren {z' \circ x}\) | \(=\) | \(\, \ds z' \circ x \, \) | \(\, \ds = \, \) | \(\ds y' \circ \paren {z' \circ x}\) | Definition of Commutative Semigroup | ||||||||
| \(\text {(4)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\, \ds y' \, \) | \(\, \ds = \, \) | \(\ds y \circ y'\) | from $(2)$ |
Thus we have:
| \(\ds y\) | \(=\) | \(\ds y' \circ y\) | from $(3)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds y \circ y'\) | Definition of Commutative Semigroup | |||||||||||
| \(\ds \) | \(=\) | \(\ds y'\) | from $(4)$ |
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.16 \ \text {(a)}$