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

$\blacksquare$


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