Inverse of Product of Subsets of Group/Proof 1
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Theorem
Let $\struct {G, \circ}$ be a group.
Let $X, Y \subseteq G$.
Then:
- $\paren {X \circ Y}^{-1} = Y^{-1} \circ X^{-1}$
where $X^{-1}$ is the inverse of $X$.
Proof
First, note that:
| \(\ds \) | \(\) | \(\ds x \in X, y \in Y\) | ||||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds x^{-1} \in X^{-1}, y^{-1} \in Y^{-1}\) | Definition of Inverse of Subset of Group | |||||||||||
| \(\ds \) | \(\leadsto\) | \(\ds y^{-1} \circ x^{-1} \in Y^{-1} \circ X^{-1}\) | Definition of Subset Product |
Now:
| \(\ds x \circ y\) | \(\in\) | \(\ds X \circ Y\) | Definition of Subset Product | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \paren {x \circ y}^{-1}\) | \(\in\) | \(\ds \paren {X \circ Y}^{-1}\) | Definition of Inverse of Subset of Group | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y^{-1} \circ x^{-1}\) | \(\in\) | \(\ds \paren {X \circ Y}^{-1}\) | Inverse of Group Product | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds Y^{-1} \circ X^{-1}\) | \(\subseteq\) | \(\ds \paren {X \circ Y}^{-1}\) | Definition of Subset |
By a similar argument we see that $\paren {X \circ Y}^{-1} \subseteq Y^{-1} \circ X^{-1}$.
Hence the result.
$\blacksquare$