# The Normalizer of a Proper Subgroup of a Nilpotent Group is Strictly Bigger

## Problem 523

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 $H \subsetneq N_G(H)$.

Let $G$ be an abelian group and let $H$ be the subset of $G$ consisting of all elements of $G$...