Product Inverse Operation Properties/Lemma 4
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Theorem
Let $\struct {G, \oplus}$ be a closed algebraic structure on which the following properties hold:
\((\text {PI} 1)\) | $:$ | Self-Inverse Property | \(\ds \forall x \in G:\) | \(\ds x \oplus x = e \) | |||||
\((\text {PI} 2)\) | $:$ | Right Identity | \(\ds \exists e \in G: \forall x \in G:\) | \(\ds x \oplus e = x \) | |||||
\((\text {PI} 3)\) | $:$ | Product Inverse with Right Identity | \(\ds \forall x, y \in G:\) | \(\ds e \oplus \paren {x \oplus y} = y \oplus x \) | |||||
\((\text {PI} 4)\) | $:$ | Cancellation Property | \(\ds \forall x, y, z \in G:\) | \(\ds \paren {x \oplus z} \oplus \paren {y \oplus z} = x \oplus y \) |
These four stipulations are known as the product inverse operation axioms.
Let $\circ$ be the operation on $G$ defined as:
- $\forall x, y \in G: x \circ y = x \oplus \paren {e \oplus y}$
Then:
- $\forall x, y, z \in G: x \oplus z = y \oplus z \implies x = y$
Proof
Let $x \oplus z = y \oplus z$.
Then we have:
\(\ds \forall x, y, z \in G: \, \) | \(\ds \paren {x \oplus z} \oplus \paren {y \oplus z}\) | \(=\) | \(\ds \paren {x \oplus z} \oplus \paren {x \oplus z}\) | from $x \oplus z = y \oplus z$ | ||||||||||
\(\ds \) | \(=\) | \(\ds x \oplus x\) | $\text {PI} 4$: Cancellation Property | |||||||||||
\(\ds \) | \(=\) | \(\ds e\) | $\text {PI} 1$: Self-Inverse Property |
Then:
\(\ds \forall x, y, z \in G: \, \) | \(\ds \paren {x \oplus z} \oplus \paren {y \oplus z}\) | \(=\) | \(\ds x \oplus y\) | $\text {PI} 4$: Cancellation Property |
Thus we have:
- $x \oplus y = e$
as both are equal to $\paren {x \oplus z} \oplus \paren {y \oplus z}$.
Then from Lemma $3$:
- $x = y$
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(b)}$