Product of Subset with Intersection/Corollary
Jump to navigation
Jump to search
Theorem
Let $\struct {G, \circ}$ be a group.
Let $X, Y, Z \subseteq G$ such that $X$ is a singleton.
Then:
- $X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$
- $\paren {Y \cap Z} \circ X = \paren {Y \circ X} \cap \paren {Z \circ X}$
where $X \circ Y$ denotes the subset product of $X$ and $Y$.
Proof
Let $X = \set x$.
We have from Product of Subset with Intersection that:
- $X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
Let $g \in \paren {X \circ Y} \cap \paren {X \circ Z}$.
Then by definition of subset product:
- $\exists y \in Y, z \in Z: g = x \circ y = x \circ z$
By the Cancellation Laws it follows that $y = z$.
But:
- $g = x \circ y \implies g \in X \circ Y$
- $g = x \circ z \implies g \in X \circ Z$
and so as $x \circ y = x \circ z$ it follows that $g \in X \circ \paren {Y \cap Z}$.
Thus:
- $X \circ \paren {Y \cap Z} \supseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
It follows by definition of set equality that:
- $X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$
By similar reasoning:
- $\paren {Y \cap Z} \circ X = \paren {Y \circ X} \cap \paren {Z \circ X}$
Hence the result.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $4 \ \text{(iii)}$