# Sufficient Condition for Quotient Group by Intersection to be Abelian

## Theorem

Let $G$ be a group.

Let $N$ and $K$ be normal subgroups of $G$.

Let the quotient groups $G / N$ and $G / K$ be abelian.

Then the quotient group $G / \paren {N \cap K}$ is also abelian.

## Proof

From Intersection of Normal Subgroups is Normal, we have that $N \cap K$ is normal in $G$.

We are given that $G / N$ and $G / K$ are abelian.

Hence:

 $\displaystyle \forall x, y \in G: \ \$ $\displaystyle \sqbrk {x, y}$ $\in$ $\displaystyle N$ Quotient Group is Abelian iff All Commutators in Divisor $\, \displaystyle \land \,$ $\displaystyle \sqbrk {x, y}$ $\in$ $\displaystyle K$ $\displaystyle \leadsto \ \$ $\displaystyle \forall x, y \in G: \ \$ $\displaystyle \sqbrk {x, y}$ $\in$ $\displaystyle N \cap K$ Definition of Set Intersection $\displaystyle \leadsto \ \$ $\displaystyle G / \paren {N \cap K}$  $\displaystyle \text {is abelian}$ Quotient Group is Abelian iff All Commutators in Divisor

$\blacksquare$