# Inverse of Product of Subsets of Group/Proof 1

## 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.

 $\displaystyle$  $\displaystyle x \in X, y \in Y$ $\displaystyle$ $\implies$ $\displaystyle x^{-1} \in X^{-1}, y^{-1} \in Y^{-1}$ 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.

$\blacksquare$