= Normalizer
{title2=$N_G(H)$}
= Normaliser
{synonym}
For a <subgroup> $H$ of a <group> $G$, the <normalizer> is
$$
N_G(H)=\{g\in G:gHg^{-1}=H\}.
$$
It is the <stabilizer subgroup> of $H$ under the <conjugation action> on <subgroups>. Its elements can permute members of $H$; they need not commute with them. For $r=(123)$ in the <symmetric group> $S_3$, the <cyclic subgroup> $H=\langle r\rangle$ has <normalizer> $S_3$, although the <centralizer> of $r$ has only three elements, as illustrated in <centralizer and normalizer>.
Every element of $H$ preserves $H$ under <conjugation>, so $H\subseteq N_G(H)$. Moreover, $H$ is a <normal subgroup> of $N_G(H)$, and this is the largest <subgroup> of $G$ with that property. Indeed, any <subgroup> in which $H$ is a <normal subgroup> consists entirely of elements preserving $H$ under <conjugation>. In particular, $N_G(H)=G$ exactly when $H$ is a <normal subgroup> of $G$.<Group Theory (v4.01) by James S. Milne>{ref}
The <orbit-stabilizer theorem> counts distinct <conjugate subsets> of a <subgroup> by
$$
\#\{gHg^{-1}:g\in G\}=[G:N_G(H)].
$$
For $H=\langle(123)\rangle$ in $S_3$, there is just one, since $[S_3:N_{S_3}(H)]=6/6=1$. For the <subgroup> $K=\langle(12)\rangle=\{e,(12)\}$, preserving $K$ forces its unique nonidentity element to stay fixed. Hence $N_{S_3}(K)=C_{S_3}((12))=K$, and its three distinct <conjugate subsets> are $\{e,(12)\}$, $\{e,(13)\}$ and $\{e,(23)\}$. This is the <subgroup> counterpart of counting a <conjugacy class> with a <centralizer>: the object being stabilized determines which denominator belongs in the count.<Group Theory (v4.01) by James S. Milne>{ref}
The equality in the definition is essential. For a <subgroup> $H$ that is a <finite group>, an inclusion $gHg^{-1}\subseteq H$ already gives equality by equal <cardinalities>. For an infinite <subgroup>, that inference can fail: <conjugation> might carry it onto a proper <subgroup>. The <normal subgroup> criterion is therefore, with the explicit hypothesis $H\leq G$,
$$
H\mathrel{\trianglelefteq}G\quad\Longleftrightarrow\quad N_G(H)=G.
$$
No finiteness or <commutativity> assumption is needed. A <normal subgroup> is preserved as a <set>; its elements need not lie in the <center of a group>.
There is a second useful count, distinct from counting conjugate <subgroups>. The <normalizer> acts on $H$ itself by <group automorphisms>, giving the <group homomorphism>
$$
\theta:N_G(H)\longrightarrow\operatorname{Aut}(H),\qquad \theta(g)(h)=ghg^{-1}.
$$
Its <group kernel> is $C_G(H)$, so the <centralizer of a subset> is a <normal subgroup> of $N_G(H)$ and the <first isomorphism theorem> gives
$$
N_G(H)/C_G(H)\cong\operatorname{im}\theta\leq\operatorname{Aut}(H).
$$
This <quotient group> records the different rearrangements of $H$ induced by the surrounding <group>. It gives a <faithful group action> on $H$, but need not be the entire <automorphism group> of $H$. For example, take the ambient <group> $G=H=\langle(123)\rangle$. Since $G$ is an <abelian group>, $N_G(H)=C_G(H)=G$, so the <quotient group> is trivial; nevertheless $r\mapsto r^{-1}$ is a nonidentity <group automorphism> of $H$. Returning to the ambient <symmetric group> $S_3$ makes that <group automorphism> available through a <reflection>, and $N_{S_3}(H)/C_{S_3}(H)\cong C_2$. Its two <cosets> are the three <rotations> and the three <reflections>.
The <subgroup> $H$ itself need not be contained in $C_G(H)$: their intersection is the <center of a group> $Z(H)$. It is contained precisely when $H$ is an <abelian group>. Thus $N_G(H)/H$, which exists because $H$ is a <normal subgroup> of its <normalizer>, and $N_G(H)/C_G(H)$ answer different questions and should not be substituted for one another.
Back to article page
