Invertible Element containing Identity in Power Structure
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Theorem
Let $\struct {S, \circ}$ be a magma.
Let identity element $e \in S$ be an identity element of $\struct {S, \circ}$.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Let $X \subseteq S$ such that:
- $e \in X$
- $X$ is invertible for $\circ_PP$.
Then $X = \set e$.
Proof
Let $X \subseteq S$ be invertible for $\circ_PP$ and such that $e \in X$.
From Identity Element for Power Structure, $\struct {\powerset S, \circ_\PP}$ has an identity element $J = \set e$.
We have:
\(\text {(1)}: \quad\) | \(\ds \exists Y \in \powerset S: \, \) | \(\ds X \circ_\PP Y\) | \(=\) | \(\ds J\) | Definition of Invertible Element: here $Y$ is the inverse of $X$ | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \set {x \circ y: x \in X, y \in Y}\) | \(=\) | \(\ds \set e\) | Definition of Operation Induced on Power Set, Definition of $J$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall y \in Y: \, \) | \(\ds e \circ y\) | \(=\) | \(\ds e\) | applying the above to $e$ as an element of $X$ | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall y \in Y: \, \) | \(\ds y\) | \(=\) | \(\ds e\) | Definition of Identity Element | |||||||||
\(\ds \leadsto \ \ \) | \(\ds Y\) | \(=\) | \(\ds \set e\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds X \circ_\PP \set e\) | \(=\) | \(\ds \set e\) | substituting for $Y$ and $J$ in $(1)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall x \in X: \, \) | \(\ds x\) | \(=\) | \(\ds e\) | Definition of Identity Element | |||||||||
\(\ds \leadsto \ \ \) | \(\ds X\) | \(=\) | \(\ds \set e\) |
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.7 \ \text {(a)}$