Intersection of Quotient Groups

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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:

\(\displaystyle \map \alpha A \cap \map \alpha B\) \(=\) \(\displaystyle \set {g N: g \in A \cap B}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {A \cap B} / N\)
\(\displaystyle \) \(=\) \(\displaystyle \map \alpha {A \cap B}\)

$\blacksquare$


Sources