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:
| \(\ds \forall x, y \in G: \, \) | \(\ds \sqbrk {x, y}\) | \(\in\) | \(\ds N\) | Quotient Group is Abelian iff All Commutators in Divisor | ||||||||||
| \(\, \ds \land \, \) | \(\ds \sqbrk {x, y}\) | \(\in\) | \(\ds K\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall x, y \in G: \, \) | \(\ds \sqbrk {x, y}\) | \(\in\) | \(\ds N \cap K\) | Definition of Set Intersection | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds G / \paren {N \cap K}\) | \(\) | \(\ds \text {is abelian}\) | 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 $16$