Product of Subset with Intersection/Equality does not Hold

Theorem

Let $\struct {G, \circ}$ be a group.

Let $X, Y, Z \subseteq G$.

Let $X \circ Y$ denotes the subset product of $X$ and $Y$.

While it is the case that:

$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$

it is not necessarily the case that:

$X \circ \paren {Y \cap Z} = \paren {X \circ Y} \cap \paren {X \circ Z}$

Proof

Proof by Counterexample:

Let $a \in G$ such that $a \ne a^{-1}$.

Let $X = \set {a, a^{-1} }, Y = \set a, Z = \set {a^{-1} }$.

Then:

$X \circ \paren {Y \cap Z} = X \circ \O = \O$

$\paren {X \circ Y} \cap \paren {X \circ Z} = \set {a^2, e} \cap \set {e, a^{-2} } \ne \O$

so:

$X \circ \paren {Y \cap Z} \ne \paren {X \circ Y} \cap \paren {X \circ Z}$

$\blacksquare$

Also see

• Product of Subset with Intersection/Corollary: equality holds when $X$ is a singleton.

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $4 \ \text{(ii)}$