# 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:

$z^{-1} \in X \circ Y$
$z^{-1} \in \paren {X^{-1} }^{-1} \circ \paren {Y^{-1} }^{-1}$

Thus by the superset proof above:

$z^{-1} \in \paren {Y^{-1} \circ X^{-1} }^{-1}$

From the definition of inverse of subset:

$z \in Y^{-1} \circ X^{-1}$

$\blacksquare$