# If Quotient $G/H$ is Abelian Group and $H < K \triangleleft G$, then $G/K$ is Abelian

## Problem 341

Let $H$ and $K$ be normal subgroups of a group $G$.
Suppose that $H < K$ and the quotient group $G/H$ is abelian.
Then prove that $G/K$ is also an abelian group.

## Solution.

We will give two proofs.

## Hint (The third isomorphism theorem)

Recall the third isomorphism theorem of groups:
Let $G$ be a group and let $H, K$ be normal subgroups of $G$ with $H < K$.
Then we have $G/K$ is a normal subgroup of $G/H$ and we have an isomorphism
$G/K \cong (G/H)/(G/K).$

## Proof 1 (Using third isomorphism theorem)

Since $H, K$ are normal subgroups of $G$ and $H < K$, the third isomorphism theorem yields that
$G/K \cong (G/H)/(G/K).$

Since the group $G/H$ is abelian by assumption, and in general a quotient group of an abelian group is abelian, it follows $(G/H)/(G/K)$ is an abelian group.

Hence by the above isomorphism, the group $G/K$ is also an abelian group.

## Proof 2 (Using the commutator subgroup)

Here is another proof using the commutator subgroup $[G, G]$ of $G$.
Recall that for a subgroup $N$ of $G$, the following two conditions are equivalent.

1. The subgroup $N$ is normal and the $G/N$ is an abelian.
2. The commutator subgroup $[G, G]$ is a subgroup of $N$.

For the proof of this fact, see the post “Commutator subgroup and abelian quotient group“.

Now we prove the problem using this fact.
Since $H$ is normal and the quotient $G/H$ is an abelian group, the commutator subgroup $[G, G]$ is a subgroup of $H$ by the fact (1 $\implies$ 2).

Then we have
\begin{align*}
[G, G] < H < K.
\end{align*}
Hence $[G, G]$ is a subgroup of $K$, hence $G/K$ is an abelian group by the fact again (2 $\implies$ 1).

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