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 $ab=ba$.

Consider the commutator of $a$ and $b$, that is, $aba^{-1}b^{-1}$.

Solution.

Consider the product $aba^{-1}b^{-1}$. Since $A$ is normal in $G$, the element $ba^{-1}b^{-1} \in A$ as it is the conjugate of $a^{-1}\in A$.
Thus $aba^{-1}b^{-1}=a(ba^{-1}b^{-1} ) \in A$.

Similarly, since $B$ is normal in $G$, we have $aba^{-1} \in B$.

Hence $aba^{-1}b^{-1}=(aba^{-1})b^{-1} \in B$.
Therefore $aba^{-1}b^{-1} \in A\cap B=\{e\}$ and we see that $aba^{-1}b^{-1}=e$, thus $ab=ba$.

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 […]

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 […]

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 […]

Prove a Group is Abelian if $(ab)^3=a^3b^3$ and No Elements of Order $3$
Let $G$ be a group. Suppose that we have
\[(ab)^3=a^3b^3\]
for any elements $a, b$ in $G$. Also suppose that $G$ has no elements of order $3$.
Then prove that $G$ is an abelian group.
Proof.
Let $a, b$ be arbitrary elements of the group $G$. We want […]

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 […]

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 […]