Inverse of Product of Subsets of Group
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 1
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.
$\blacksquare$
Proof 2
Superset
We will show that $\forall z \in Y^{-1} \circ X^{-1}: z \in \paren {X \circ Y}^{-1}$, from which:
- $Y^{-1} \circ X^{-1} \subseteq \paren {X \circ Y}^{-1}$
Let $z \in Y^{-1} \circ X^{-1}$.
By the definition of subset product:
- $\exists x' \in X^{-1}, y' \in Y^{-1}: z = y' \circ x'$
Then by Inverse of Group Product:
- $(2)\quad z^{-1} = x'^{-1} \circ y'^{-1}$
By the definition of inverse of subset:
- $x'^{-1} \in X$ and $y'^{-1} \in Y$
By the definition of subset product:
- $x'^{-1} \circ y'^{-1} \in X \circ Y$
Thus by $(2)$:
- $z^{-1} \in X \circ Y$
By the definition of inverse of subset:
- $z \in \paren {X \circ Y}^{-1}$
Subset
We will show that $\forall z \in \paren {X \circ Y}^{-1}: z \in Y^{-1} \circ X^{-1}$, from which:
- $\paren {X \circ Y}^{-1} \subseteq Y^{-1} \circ X^{-1}$
Let $z \in \paren {X \circ Y}^{-1}$.
By the definition of inverse of subset:
- $z^{-1} \in X \circ Y$
From Inverse of Inverse of Subset of Group:
- $z^{-1} \in \paren {X^{-1} }^{-1} \circ \paren {Y^{-1} }^{-1}$
Thus by the superset proof above:
- $z^{-1} \in \paren {Y^{-1} \circ X^{-1} }^{-1}$
From the definition of inverse of subset:
- $z \in Y^{-1} \circ X^{-1}$
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Exercise $\text{G}$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $6 \ \text{(i)}$