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 finite group and $P$ be a nontrivial Sylow subgroup of $G$.
Let $H$ be a subgroup of $G$ containing the normalizer $N_G(P)$ of $P$ in $G$.

Then show that $N_G(H)=H$.
 

Hint.

Use the conjugate part of the Sylow theorem.
See the second statement of the Sylow theorem.

Proof.

It is clear that $H \subset N_G(H)$.
So we show that $N_G(H) \subset H$.

Take any $a \in N_G(H)$. Since $P \subset N_G(P) \subset H$, we have
\[aPa^{-1} \subset aHa^{-1}=H,\] where the last step follows from $a \in N_G(H)$.

It follows that $P$ and $aPa^{-1}$ are both Sylow subgroups in $H$.
By the Sylow theorem, any two $p$-Sylow subgroups are conjugate. Thus there exists $b \in H$ such that $bPb^{-1}=aPa^{-1}$.

This implies that $(b^{-1}a)P(b^{-1}a)^{-1}=P$ and thus $b^{-1}a \in N_G(P) \subset H$. Hence we have $a \in H$ since $b \in H$.
This shows that $N_G(H) \subset H$, hence $N_G(H)=H$ as required.

Corollary (The Normalizer of the Normalizer of a Sylow subgroup)

We apply the result to the case $H=N_G(P)$, and obtain the following result.

The post If a Subgroup Contains a Sylow Subgroup, then the Normalizer is the Subgroup itself first appeared on Problems in Mathematics.

Let $H$ be a subgroup of order $2$. Let $N_G(H)$ be the normalizer of $H$ in $G$ and $C_G(H)$ be the centralizer of $H$ in $G$.

(a) Show that $N_G(H)=C_G(H)$.

(b) If $H$ is a normal subgroup of $G$, then show that $H$ is a subgroup of the center $Z(G)$ of $G$.

Definitions.

Recall that the centralizer of $H$ in $G$ is
\[C_G(H)=\{g \in G \mid gh=hg \text{ for any } h\in H\}.\]

The normalizer of $H$ in $G$ is
\[N_G(H)=\{g\in G \mid gH=Hg \}.\]

Proof.

(a) Prove $N_G(H)=C_G(H)$

In general, we have $C_G(H) \subset N_G(H)$. We show that $N_G(H) \subset C_G(H)$.

Take any $g \in N_G(H)$. We have $gH=Hg$. Since $|H|=2$, let $H=\{1,h\}$.
Then $gH=\{g,gh\}$ and $Hg=\{g, hg\}$. Since $gH=Hg$, we have $gh=hg$. Namely $g\in C_G(H)$.

This proves that $N_G(H) \subset C_G(H)$, hence $N_G(H)= C_G(H)$.

(b) If $H$ is normal, then $H$ is a subgroup of $Z(G)$

Suppose that $H$ is a normal subgroup of $G$, that is $G=N_G(H)$.
By part (a), this implies that $G=C_G(H)$. Hence $H < Z(G)$.

The post Normalizer and Centralizer of a Subgroup of Order 2 first appeared on Problems in Mathematics.

Let $D_8$ be the dihedral group of order $8$.
Using the generators and relations, we have
\[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle.\]

(a) Let $A$ be the subgroup of $D_8$ generated by $r$, that is, $A=\{1,r,r^2,r^3\}$.
Prove that the centralizer $C_{D_8}(A)=A$.

(b) Show that the normalizer $N_{D_8}(A)=D_8$.

(c) Show that the center $Z(D_8)=\langle r^2 \rangle=\{1,r^2\}$, the subgroup generated by $r^2$.

Definitions (centralizer, normalizer, center).

Recall the definitions.

1.  The centralizer $C_{D_8}(A)$ is a subgroup of $D_8$ whose elements commute with $A$.
   That is $C_{D_8}(A)=\{ g\in D_8 \mid gxg^{-1}=x \text{ for all } x\in A\}$.
2. The normalizer $N_{D_8}(A)$ is a subgroup of $D_8$ defined as
   $N_{D_8}(A)=\{ g\in D_8 \mid gxg^{-1} \in A \text{ for any } x \in A\}$.
3. The center $Z(D_8)$ is a subgroup of $D_8$ whose elements commute with all elements of $D_8$.
   That is, $Z(D_8)=\{g \in D_8 \mid gxg^{-1}=x \text{ for all } x\in D_8\}$.

Proof.

(a) The centralizer $C_{D_8}(A)=A$

 Since any power of $r$ commutes with each other we have $A < C_{D_8}(A)$.
Since $sr=r^{-1}s$ and $r^{-1}s\neq rs$ (otherwise $r^2=1$), we see that $s\not \in C_{D_8}(A)$.

This also implies that any element of the form $r^as \in D_8$ is not in $C_{D_8}(A)$. In fact, if $r^as\in C_{D_8}(A)$, then $s=(r^{-a})\cdot (r^a s)$ is also in $C_{D_8}(A)$ because $r^{-a}\in C_{D_8}(A)$ and $C_{D_8}(A)$ is a group.

This is a contradiction. Therefore $C_{D_8}(A)=A$.

(b) The normalizer $N_{D_8}(A)=D_8$

 In general, the centralizer of a subset is contained in the normalizer of the subset. From this fact we have $A=C_{D_8}(A) < N_{D_8}(A)$.
Thus it suffices to show that the other generator $s \in D_8$ belongs to $N_{D_8}(A)$.

We have $sr^as^{-1}=r^{-1}ss^{-1}=r^{-1}\in A$ using the relation $sr=r^{-1}s$.
Thus $s \in N_{D_8}(A)$ as required.

(c) The center $Z(D_8)=\langle r^2 \rangle=\{1,r^2\}$

The center $Z(D_8)$ is contained in the centralizer $C_{D_8}(A)=A$.
Since $rsr^{-1}=sr^2 \neq s$ and $r^3s(r^3)^{-1}=r^{-1}sr=sr^2\neq s$, the elements $r$ and $r^3$ are not in the center $Z(D_8)$.

On the other hand, we have $sr^2=r^{-1}sr=r^{-2}s=r^2s$. Thus $r^2s(r^2)^{-1}=s$ and $r^2 \in Z(D_8)$.
Therefore we have $Z(D_8)=\{1, r^2\}$.

The post Centralizer, Normalizer, and Center of the Dihedral Group $D_{8}$ first appeared on Problems in Mathematics.