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 $G$.

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 $G$.
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 $N_G(H)=H$.
Read solution
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 $G$.
Read solution
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 $r^2$.