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 $N \supset D(G)$.
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.
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 […]
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Definitions/Hint.
We first recall relevant definitions.
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Let $H$ be a normal subgroup of a group $G$.
Then show that $N:=[H, G]$ is a subgroup of $H$ and $N \triangleleft G$.
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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 $[h, k]:=hkh^{-1}k^{-1}$.
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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.
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Quotient Group of Abelian Group is Abelian
Let $G$ be an abelian group and let $N$ be a normal subgroup of $G$.
Then prove that the quotient group $G/N$ is also an abelian group.
Proof.
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