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The symmetric group $S_3$ has exactly six permutations, written using permutation cycles:Here $e$ is the identity element; $(12)$, $(13)$ and $(23)$ are transpositions; and $(123)$ and $(132)$ are permutation cycles of length three. With $t=(12)$ and $r=(123)$, rightmost-first function composition gives $r^2=(132)$ and $r^{-1}=r^2$. The conjugacy classes and representative centralizers are

An explicit check of the conjugation $gt{g}^{-1}$ with $g=r=(123)$ sends $1\mapsto 3\mapsto 3\mapsto 1$, $2\mapsto 1\mapsto 2\mapsto 3$, and $3\mapsto 2\mapsto 1\mapsto 2$, where each chain applies $g^{-1}$, then $t$, then $g$. Thus $1$ is fixed and $2,3$ are exchanged: $gt{g}^{-1}=(23)$. Relabelling also produces $(13)$, so the three transpositions share one conjugacy class. A transposition reverses the length-three permutation cycle, since $tr{t}^{-1}=(213)=(132)=r^{-1}$, putting both length-three permutation cycles in one conjugacy class.

Checking each of the six possible permutations against $gx=xg$ gives the displayed centralizers. Only $e$ has a one-element conjugacy class, so the center of a group here is $Z(S_3)=\{e\}$. The class equation is $6=1+3+2$. The second diagram in centralizer and normalizer uses the same $r$ and $t$ to show how preserving a cyclic subgroup differs from fixing its group generator.

For a subgroup $H$ of a group $G$, the normalizer isIt 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 $H=⟨r⟩$ 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 $G$.^[ref]

The orbit-stabilizer theorem counts distinct conjugate subsets of a subgroup byFor $H=⟨(123)⟩$ in $S_3$, there is just one, since $[S_3:N_{S_3}(H)]=6/6=1$. For the subgroup $K=⟨(12)⟩=\{e,(12)\}$, 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.^[ref]

The equality in the definition is essential. For a subgroup $H$ that is a finite group, an inclusion $gH{g}^{-1}\subseteq H$ 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 $H\le G$,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 homomorphismIts 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 givesThis 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 $G=H=⟨(123)⟩$. 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.