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

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

Then:

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

So we have:

 $\displaystyle y$ $\in$ $\displaystyle X$ $\, \displaystyle \land \,$ $\displaystyle s$ $\in$ $\displaystyle X$ Definition of Subset $\displaystyle \leadsto \ \$ $\displaystyle y^{-1} \circ s$ $\in$ $\displaystyle X$ One-Step Subgroup Test $\displaystyle \leadsto \ \$ $\displaystyle z$ $\in$ $\displaystyle X$ 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 $X \cap \paren {Y \circ Z}$

 $\displaystyle Y \circ X$ $\subseteq$ $\displaystyle X \circ X$ Subset of Subset Product $(1):\quad$ $\displaystyle \leadsto \ \$ $\displaystyle Y \circ X$ $\subseteq$ $\displaystyle X$ Product of Subgroup with Itself

Then:

 $\displaystyle Y \circ \paren {X \cap Z}$ $\subseteq$ $\displaystyle \paren {Y \circ X} \cap \paren {Y \circ Z}$ Product of Subset with Intersection $\displaystyle$ $\subseteq$ $\displaystyle X \cap \paren {Y \circ Z}$ 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:

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

$\blacksquare$