Conditions under which Commutative Semigroup is Group/Lemma 2
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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 $y \circ x = x$, then $y \circ y = y$.
Proof
\(\ds y \circ x\) | \(=\) | \(\ds x\) | by hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y \circ \paren {y \circ x}\) | \(=\) | \(\ds y \circ x\) | premultiplying both sides by $y$ | ||||||||||
\(\ds \) | \(=\) | \(\ds x\) | by hypothesis | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {y \circ y} \circ x\) | \(=\) | \(\ds x\) | Semigroup Axiom $\text S 1$: Associativity | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y \circ y\) | \(=\) | \(\ds y\) | Lemma 1 |
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.16 \ \text {(b)}$