# Commutator Subgroup and Abelian Quotient Group

## Problem 147

Let $G$ be a group and let $D(G)=[G,G]$ be the commutator subgroup of $G$.
Let $N$ be a subgroup of $G$.
Prove that the subgroup $N$ is normal in $G$ and $G/N$ is an abelian group if and only if $N \supset D(G)$.

## Definitions.

Recall that for any $a, b \in G$, the commutator of $a$ and $b$ is
$[a,b]=a^{-1}b^{-1}ab \in G.$ The commutator subgroup $D(G)=[G,G]$ is a subgroup of $G$ generated by all commutators.
That is,
$D(G)=[G,G]=\langle [a,b] \mid a,b \in G \rangle.$

## Proof.

$(\implies)$ Suppose that $N$ is a normal subgroup of $G$ and the quotient $G/N$ is an abelian group.
Then for any elements $a, b \in G$, we have
\begin{align*}
abN=aN\cdot bN=bN \cdot aN=baN.
\end{align*}
(Here we used the fact that $N$ is normal, hence $G/N$ is a group.)

From this, we obtain that
$a^{-1}b^{-1}abN=N$ and thus $[a,b]=a^{-1}b^{-1}ab\in N$.
Since any generator $[a,b]$ is in $N$, we have $D(G)\subset N$.

$(\impliedby)$ On the other hand, let us assume that $N \supset D(G)$.
We first show that $N$ is a normal subgroup of $G$.
For any $g \in G$, $x\in N$, we have
\begin{align*}
gxg^{-1}=gxg^{-1}x^{-1}x=[g^{-1},x^{-1}]\cdot x\in N
\end{align*}
since the commutator $[g^{-1},x^{-1}]\in D(G)\subset N$ and $x \in N$.
Thus $N$ is normal in $G$.

Now that $N$ is normal in $G$, the quotient $G/N$ is a group. We show that $G/N$ is an abelian group.
For any $a, b \in G$, we have
\begin{align*}
aN\cdot bN&=ab N\\
&=baa^{-1}b^{-1}abN\\
&=ba[a,b]N\\
&=baN \qquad \text{ since } [a,b] \in N\\
&=bN\cdot aN.
\end{align*}
Therefore the group operations of $G/N$ is commutative, and hence $G/H$ is abelian.

## Related Question.

For another abelian group problem, check out
Two quotients groups are abelian then intersection quotient is abelian

• Two Quotients Groups are Abelian then Intersection Quotient is Abelian 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 […]
• Non-Abelian Simple Group is Equal to its Commutator Subgroup Let $G$ be a non-abelian simple group. Let $D(G)=[G,G]$ be the commutator subgroup of $G$. Show that $G=D(G)$.   Definitions/Hint. We first recall relevant definitions. A group is called simple if its normal subgroups are either the trivial subgroup or the group […]
• A Condition that a Commutator Group is a Normal Subgroup Let $H$ be a normal subgroup of a group $G$. Then show that $N:=[H, G]$ is a subgroup of $H$ and $N \triangleleft G$. Here $[H, G]$ is a subgroup of $G$ generated by commutators $[h,k]:=hkh^{-1}k^{-1}$. In particular, the commutator subgroup $[G, G]$ is a normal subgroup of […]
• Group Generated by Commutators of Two Normal Subgroups is a Normal Subgroup Let $G$ be a group and $H$ and $K$ be subgroups of $G$. For $h \in H$, and $k \in K$, we define the commutator $[h, k]:=hkh^{-1}k^{-1}$. Let $[H,K]$ be a subgroup of $G$ generated by all such commutators. Show that if $H$ and $K$ are normal subgroups of $G$, then the subgroup […]
• If a Subgroup $H$ is in the Center of a Group $G$ and $G/H$ is Nilpotent, then $G$ is Nilpotent Let $G$ be a nilpotent group and let $H$ be a subgroup such that $H$ is a subgroup in the center $Z(G)$ of $G$. Suppose that the quotient $G/H$ is nilpotent. Then show that $G$ is also nilpotent.   Definition (Nilpotent Group) We recall here the definition of a […]
• Normal Subgroups, Isomorphic Quotients, But Not Isomorphic Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$. Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.   Proof. We give a […]
• A Simple Abelian Group if and only if the Order is a Prime Number Let $G$ be a group. (Do not assume that $G$ is a finite group.) Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.   Definition. A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is […]