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:

\(\ds \map \alpha A \cap \map \alpha B\)

\(=\)

\(\ds \set {g N: g \in A \cap B}\)

\(\ds \)

\(=\)

\(\ds \paren {A \cap B} / N\)

\(\ds \)

\(=\)

\(\ds \map \alpha {A \cap B}\)

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Proposition $7.16 \ \text{(i)}$