# 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 $G$. Add to solve later

## 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 $H \subsetneq N_G(H)$. Add to solve later

### 2 Responses

1. Alka says:

In problem 523,I am having a doubt regarding the inclusion of G”k+1 in H.Is it confirmed that any subgroup of G can be placed between two consecutive elements in its lower central series?thankyou.

• Yu says:

Dear Alka,

The statement
$G^{k+1} \subset H \text{ but } G^{k} \nsubseteq H$
does not say $H$ is placed between two consecutive elements in its lower central series. Consider the lower central series from the identity element $G^n$. Of course, the identity element is contained in $H$. Is the next component $G^{n-1}$ contained in $H$? If so, how about next? Eventually, there is some $k$ such that $G^k$ is not contained in $H$ as $H$ is a proper subgroup.

I hope this helps.

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