# Intersection with Subgroup Product of Superset

## Theorem

Let $X, Y, Z$ be subgroups of a group $\struct {G, \circ}$.

Let $Y \subseteq X$.

Then:

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

where $Y \circ Z$ denotes subset product.

## Proof

By definition of set equality, it suffices to prove:

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

and:

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

### $X \cap \paren {Y \circ Z}$ is contained in $Y \circ \paren {X \cap Z}$

Let $s \in X \cap \paren {Y \circ Z}$.

 $\ds s$ $\in$ $\ds X \cap \paren {Y \circ Z}$ $\ds \leadsto \ \$ $\ds s$ $\in$ $\ds X$ Definition of Set Intersection $\, \ds \land \,$ $\ds s$ $\in$ $\ds Y \circ Z$ Definition of Set Intersection $\ds \leadsto \ \$ $\ds \exists y \in Y, z \in Z: \,$ $\ds s$ $=$ $\ds y \circ z$ Definition of Subset Product $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds z$ $=$ $\ds y^{-1} \circ s$ Group has Latin Square Property

Then:

 $\ds Y$ $\subseteq$ $\ds X$ $\ds \leadsto \ \$ $\ds y$ $\in$ $\ds X$ Definition of Subset

So we have:

 $\ds y$ $\in$ $\ds X$ $\, \ds \land \,$ $\ds s$ $\in$ $\ds X$ Definition of Subset $\ds \leadsto \ \$ $\ds y^{-1} \circ s$ $\in$ $\ds X$ One-Step Subgroup Test $\ds \leadsto \ \$ $\ds z$ $\in$ $\ds X$ from $(1)$

Thus by definition of set intersection:

$z \in X \cap Z$

So:

$s = y \circ z \in Y \circ \paren {X \cap Z}$

By definition of subset:

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

$\Box$

### $Y \circ \paren {X \cap Z}$ is contained in $X \cap \paren {Y \circ Z}$

 $\ds Y \circ X$ $\subseteq$ $\ds X \circ X$ Subset Relation is Compatible with Subset Product $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds Y \circ X$ $\subseteq$ $\ds X$ Product of Subgroup with Itself

Then:

 $\ds Y \circ \paren {X \cap Z}$ $\subseteq$ $\ds \paren {Y \circ X} \cap \paren {Y \circ Z}$ Product of Subset with Intersection $\ds$ $\subseteq$ $\ds X \cap \paren {Y \circ Z}$ from $(1)$

$\Box$

We have established that:

$x \in Y \circ \paren {X \cap Z} \iff x \in X \cap \paren {Y \circ Z}$

From the definition of set equality:

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

$\blacksquare$