Inverse of Product of Subsets of Group/Proof 2
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
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:
- $z \in Y^{-1} \circ X^{-1}$
$\blacksquare$