# Two Quotients Groups are Abelian then Intersection Quotient is Abelian

## Problem 148

Let $K, N$ be normal subgroups of a group $G$. Suppose that the quotient groups $G/K$ and $G/N$ are both abelian groups.

Then show that the group
$G/(K \cap N)$ is also an abelian group.

## Hint.

We use the following fact to prove the problem.

Lemma: For a subgroup $H$ of a group $G$, $H$ is normal in $G$ and $G/H$ is an abelian group if and only if the commutator subgroup $D(G)=[G,G]$ of $G$ is contained in $H$.

For a proof of this fact, see Commutator subgroup and abelian quotient group

## Proof.

By the lemma mentioned above, we know that $G/K$ is an abelian group if and only if the commutator subgroup $D(G)=[G,G]$ is contained in $K$.

Similarly, since $G/N$ is abelian, $D(G)$ is contained in $N$.
Therefore, the commutator subgroup $D(G) \subset K \cap N$. This implies, again by Lemma, that the quotient group
$G/(K \cap N)$ is an abelian group as required.

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