Inverse Subset Group Product

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Theorem

Let $\left({G, \circ}\right)$ be a group.

Let $X, Y \subseteq G$.


Then:

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

where $X^{-1}$ is the inverse of $X$.


Proof

First, note that.

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


Now:

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

$\blacksquare$


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