# Intersection of Quotient Groups

## Theorem

Let $N \lhd G$ be a normal subgroup of $G$.

Let:

$N \le A \le G$
$N \le B \le G$

For a subgroup $H$ of $G$, let $\alpha$ be the bijection defined as:

$\map \alpha H = \set {h N: h \in H}$

Then:

$\map \alpha {A \cap B} = \map \alpha A \cap \map \alpha B$

## Proof

From the proof of the Correspondence Theorem:

$\map \alpha H \subseteq G / N$

Then:

 $\ds \map \alpha A \cap \map \alpha B$ $=$ $\ds \set {g N: g \in A \cap B}$ $\ds$ $=$ $\ds \paren {A \cap B} / N$ $\ds$ $=$ $\ds \map \alpha {A \cap B}$

$\blacksquare$