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.

We recall here the definition of a nilpotent group.
Let $G$ be group. Define $G^0=G$,
\[ G^1=[G, G]=\langle [x,y]:=xyx^{-1}y^{-1} \mid x, y \in G\rangle,\]
and inductively define
\[G^i=[G^{i-1},G]=\langle [x,y] \mid x \in G^{i-1}, y \in G \rangle.\]
Then we obtain so called the lower central series of $G$:
\[ G^{0} \triangleright G^{1} \triangleright \cdots \triangleright G^{i} \triangleright \cdots. \]

If there exists $m\in \Z$ such that $G^m=\{e\}$, then the group $G$ is called nilpotent.

Proof.

Consider the natural projection $p:G \to G/H$.
Then we have $p(G^i)=(G/H)^i$.

Since $G/H$ is nilpotent, there exists $m \in \Z$ such that $(G/H)^m=\{eH\}$.
Thus we obtain
\[p(G^m)=(G/H)^m=\{eH\}.\]
Thus for any $g \in G^m$, $g \in H \subset Z(G)$.

It follows that for any $g \in G^m$, $x \in G$ we have $gxg^{-1}x^{-1}=e$.
Since the elements $gxg^{-1}x^{-1}$ are generators of $G^{m+1}=[G^m, G]$, we conclude that $G^{m+1}=\{e\}$ and $G$ is nilpotent.

The Normalizer of a Proper Subgroup of a Nilpotent Group is Strictly Bigger
Let $G$ be a nilpotent group and let $H$ be a proper subgroup of $G$.
Then prove that $H \subsetneq N_G(H)$, where $N_G(H)$ is the normalizer of $H$ in $G$.
Proof.
Note that we always have $H \subset N_G(H)$.
Hence our goal is to find an element in […]

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 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 […]

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, 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 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 […]

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.
Each element of $G/N$ is a coset $aN$ for some $a\in G$.
Let $aN, bN$ be arbitrary elements of $G/N$, where $a, b\in […]

Infinite Cyclic Groups Do Not Have Composition Series
Let $G$ be an infinite cyclic group. Then show that $G$ does not have a composition series.
Proof.
Let $G=\langle a \rangle$ and suppose that $G$ has a composition series
\[G=G_0\rhd G_1 \rhd \cdots G_{m-1} \rhd G_m=\{e\},\]
where $e$ is the identity element of […]