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 $[H, K]$ is normal in $G$.

We first prove that a conjugate of each generator is in $[H,K]$.
Let $h \in H, k \in K$. For any $g \in G$, we have by inserting $g^{-1}g=e$ inbetweens
\begin{align*}
g[h,k]g^{-1}&=ghkh^{-1}k^{-1}=(ghg^{-1})(gkg^{-1})(gh^{-1}g^{-1})(gk^{-1}g^{-1})\\
& = (ghg^{-1})(gkg^{-1})(ghg^{-1})^{-1}(gkg^{-1})^{-1}.
\end{align*}

Now note that $ghg^{-1} \in H$ since $H$ is normal in $G$, and $gkg^{-1} \in K$ since $K$ is normal in $G$. Thus $g[h,k]g^{-1} \in [H, K]$.
By taking the inverse of the above equality, we also see that $g[k,h]g^{-1} \in [H, K]$. Thus the conjugate of the inverse $[h,k]^{-1}=[k,h]$ is in $[H, K]$.

Next, note that any element $x \in [H,K]$ is a product of generators or their inverses. So let us write
\[x=[h_1, k_1]^{\pm 1}[h_2, k_2]^{\pm 1}\cdots [h_n, k_n]^{\pm 1},\]
where $h_i \in H, k_i\in K$ for $i=1,\dots, n$.
Then for any $g \in G$, we have
\begin{align*}
gxg^{-1}=(g[h_1, k_1]^{\pm 1}g^{-1})(g[h_2, k_2]^{\pm 1}g^{-1})\cdots(g [h_n, k_n]^{\pm 1} g^{-1}).
\end{align*}

We saw that the conjugate of a generator, or its inverse, by $g \in G$ is in $[H,K]$.
Thus $gxg^{-1}$ is also in $[H, K]$.
This proves that the group $[H,K]$ 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 […]

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 […]

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 […]

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 […]

Two Normal Subgroups Intersecting Trivially Commute Each Other
Let $G$ be a group. Assume that $H$ and $K$ are both normal subgroups of $G$ and $H \cap K=1$. Then for any elements $h \in H$ and $k\in K$, show that $hk=kh$.
Proof.
It suffices to show that $h^{-1}k^{-1}hk \in H \cap K$.
In fact, if this it true then we have […]

Abelian Normal Subgroup, Intersection, and Product of Groups
Let $G$ be a group and let $A$ be an abelian subgroup of $G$ with $A \triangleleft G$.
(That is, $A$ is a normal subgroup of $G$.)
If $B$ is any subgroup of $G$, then show that
\[A \cap B \triangleleft AB.\]
Proof.
First of all, since $A \triangleleft G$, the […]

If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup
Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.
Hint.
It follows from […]

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