# Sufficient Condition for Quotient Group by Intersection to be Abelian

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## 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 |

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $16$