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 $N_G(H)$ that does not belong to $H$.

Since $G$ is a nilpotent group, it has a lower central series
\[ G=G^{0} \triangleright G^{1} \triangleright \cdots \triangleright G^{n}=\{e\},\] where $G=G^{0}$ and $G^{i}$ is defined by
\[G^i=[G^{i-1},G]=\langle [x,y]=xyx^{-1}y^{-1} \mid x \in G^{i-1}, y \in G \rangle\] successively, and $e$ is the identity element of $G$.

Since $H$ is a proper subgroup of $G$, there is an index $k$ such that
\[G^{k+1} \subset H \text{ but } G^{k} \nsubseteq H.\]

Take any $x\in G^{k} \setminus H$.
We claim that $x \in N_G(H)$.

For any $y\in H$, it follows from the definition of $G^{k+1}$ that
\[ [x,y] \in G^{k+1} \subset H.\] Hence $xyx^{-1}y^{-1}\in H$.
Since $y\in H$, we see that $xyx^{-1}\in H$.
As this is true for any $y\in H$, we conclude that $x\in N_G(H)$.
The claim is proved.

Since $x$ does not belong to $H$, we conclude that $H \subsetneq N_G(H)$.

The post The Normalizer of a Proper Subgroup of a Nilpotent Group is Strictly Bigger first appeared on Problems in Mathematics.

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.

Then show that $G$ is also nilpotent.

Definition (Nilpotent Group)

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 post If a Subgroup $H$ is in the Center of a Group $G$ and $G/H$ is Nilpotent, then $G$ is Nilpotent first appeared on Problems in Mathematics.