Intersection with Subgroup Product of Superset

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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:

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

$\blacksquare$


Sources

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