Inverse of Product of Subsets of Group

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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 1

First, note that.

\(\displaystyle \) \(\) \(\displaystyle x \in X, y \in Y\)
\(\displaystyle \) \(\implies\) \(\displaystyle x^{-1} \in X^{-1}, y^{-1} \in Y^{-1}\) Definition of Inverse of Subset of Group
\(\displaystyle \) \(\implies\) \(\displaystyle y^{-1} \circ x^{-1} \in Y^{-1} \circ X^{-1}\) Definition of Subset Product


Now:

\(\displaystyle x \circ y\) \(\in\) \(\displaystyle X \circ Y\) Definition of Subset Product
\(\displaystyle \implies \ \ \) \(\displaystyle \left({x \circ y}\right)^{-1}\) \(\in\) \(\displaystyle \left({X \circ Y}\right)^{-1}\) Definition of Inverse of Subset of Group
\(\displaystyle \implies \ \ \) \(\displaystyle y^{-1} \circ x^{-1}\) \(\in\) \(\displaystyle \left({X \circ Y}\right)^{-1}\) Inverse of Group Product
\(\displaystyle \implies \ \ \) \(\displaystyle Y^{-1} \circ X^{-1}\) \(\subseteq\) \(\displaystyle \left({X \circ Y}\right)^{-1}\) Definition of Subset


By a similar argument we see that $\left({X \circ Y}\right)^{-1} \subseteq Y^{-1} \circ X^{-1}$.


Hence the result.

$\blacksquare$


Proof 2

Superset

We will show that $\forall z \in Y^{-1} \circ X^{-1}: z \in \left({X \circ Y}\right)^{-1}$, from which:

$Y^{-1} \circ X^{-1} \subseteq \left({X \circ Y}\right)^{-1}$


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:

$z \in \left({X \circ Y}\right)^{-1}$


Subset

We will show that $\forall z \in \left({X \circ Y}\right)^{-1}: z \in Y^{-1} \circ X^{-1}$, from which:

$\left({X \circ Y}\right)^{-1} \subseteq Y^{-1} \circ X^{-1}$


Let $z \in \left({X \circ Y}\right)^{-1}$.

By the definition of inverse of subset:

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

From Inverse of Inverse of Subset of Group:

$z^{-1} \in \left({ X^{-1} }\right)^{-1} \circ \left({ Y^{-1} }\right)^{-1}$

Thus by the superset proof above:

$z^{-1} \in \left({ Y^{-1} \circ X^{-1} }\right)^{-1}$

From the definition of inverse of subset:

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

$\blacksquare$


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