First, we show that $N=[H, G]$ is a subgroup of $H$.
A generator of $N$ is of the form either $hgh^{-1} g^{-1}$ or $ghg^{-1} h^{-1}$, where $h \in H$ and $g \in G$. Since $H$ is normal in $G$, we see that these are elements in $H$. Thus $N <H$.

Next, we show that $N$ is normal in $G$.

Let $x=hgh^{-1} g^{-1}$ be a generator element of $N$.

Then for any $a \in G$, we have
\[ ax a^{-1} =ahgh^{-1} g^{-1}a^{-1}=(aha^{-1})(aga^{-1})(ah^{-1}a^{-1})(a g^{-1}a^{-1}) \in [H, G]. \]
Similarly, if $x$ is a generator of the form $ghg^{-1} h^{-1}$, then we see that $ghg^{-1} h^{-1} \in [H, G]$ by the same argument.
Thus the conjugate of a generator of $N$ by $a \in G$ stays in $N$.

Now any element $x \in N$ is of the form $x=x_1 x_2 \cdots x_n$, where $x_i$ are generators of $N$.

For any $a \in H$, we have
\[ axa^{-1}=a x_1 x_2 \cdots x_n a^{-1}= (ax_1 a^{-1})(a x_2 a^{-1}) \cdots (a x_n a^{-1}) \in [H, G].\]
Thus $N \triangleleft G$.

In particular, we apply the result to $H=G$. Then we see that the commutator subgroup $[G, G]$ is a normal subgroup of $G$.

Commutator Subgroup and Abelian Quotient Group
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 […]

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 […]

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 […]

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 […]

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 […]

Normal Subgroups Intersecting Trivially Commute in a Group
Let $A$ and $B$ be normal subgroups of a group $G$. Suppose $A\cap B=\{e\}$, where $e$ is the unit element of the group $G$.
Show that for any $a \in A$ and $b \in B$ we have $ab=ba$.
Hint.
Consider the commutator of $a$ and $b$, that […]

Prove that a Group of Order 217 is Cyclic and Find the Number of Generators
Let $G$ be a finite group of order $217$.
(a) Prove that $G$ is a cyclic group.
(b) Determine the number of generators of the group $G$.
Sylow's Theorem
We will use Sylow's theorem to prove part (a).
For a review of Sylow's theorem, check out the […]

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[…] Another problem about a commutator group is A condition that a commutator group is a normal subgroup […]

[…] The commutator subgroup $D(G)=[G,G]$ is a normal subgroup of $G$. For a proof, see: A condition that a commutator group is a normal subgroup. […]