Inverse Subset Group Product
From ProofWiki
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.
$\blacksquare$
Sources
- George McCarty: Topology: An Introduction with Application to Topological Groups (1967): Chapter $\text{II}$: Exercise $\text{G}$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $7.6 \ \text{(i)}$