If a Subgroup $H$ is in the Center of a Group $G$ and $G/H$ is Nilpotent, then $G$ is Nilpotent

Problem 128

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.

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Proof.
Note that we always have $H \subset N_G(H)$.
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Then prove that the quotient group $G/N$ is also an abelian group.
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