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Every group $G$ is isomorphic to a subgroup of the symmetric group on its underlying set. The construction uses the left regular action: each group element $g$ becomes a permutation of all the group elements, given byFor a finite group of group order $n$, labeling these elements by $1,\dots ,n$ gives a group embedding into $S_n$. The same construction works for an infinite group, with the symmetric group of all bijections of its underlying set.^[ref]

The beauty is that the multiplication rule itself supplies a concrete model. Whatever the group elements originally represent, each one rearranges the entire group by multiplication. A product becomes function composition, and no distinctions between group elements are lost. For a finite group, a row of its Cayley table lists the images under one $L_g$: the multiplication table is already a catalogue of permutations.

Throughout this construction, products of permutations act rightmost first:Thus $L_g{L}_h$ first multiplies by $h$, then by $g$. This is a left group action, not a license to reverse the multiplication order in a nonabelian group. Ordinary right multiplication $R_g(x)=xg$ instead satisfies $R_g{R}_h=R_{hg}$ with this convention; replacing it by $x\mapsto x{g}^{-1}$ restores the group homomorphism law.^[ref]

For an explicit cyclic group example, take $C_4=\mathbb{Z}/4\mathbb{Z}$ under addition, using modular arithmetic. Its left regular action consists of $L_k(x)=k+x(\mathrm{mod}4)$. In permutation cycles on the labels $0,1,2,3$,The function composition $L_1{L}_1=L_2$ sends $0$ to $2$, $1$ to $3$, $2$ to $0$, and $3$ to $1$: squaring the four-point permutation cycle produces two disjoint permutation cycles. Also $L_1{L}_3=L_0$. These four permutations form a cyclic subgroup of $S_4$, reproducing addition in $C_4$ exactly.

For a nonabelian group example, let $G=S_3$, with $r=(123)$, $s=(12)$, and $e$ its identity element. The six group elements are $e,r,r^2,s,rs,r^{2}s$, where $r^3=s^2=e$ and $sr=r^{2}s$. Now the labels being permuted are themselves permutations of three points. Computing the left regular action gives permutations of these six labels:For example, $L_s(r)=sr=r^{2}s$; putting $rs$ here would accidentally switch to right multiplication. Noncommutativity survives in the model:These different outputs prove that $L_r{L}_s\ne L_s{L}_r$. The group generators $r,s$ become group generators $L_r,L_s$ of a six-element subgroup of $S_6$.

Why does a group action here produce a group embedding? Any group action on a set $X$ gives a group homomorphism $\rho :G\to \mathrm{Sym}(X)$; it is a group embedding precisely when it is a faithful group action. The group kernel consists of the group elements fixing every point of $X$. In the left regular action, $L_g(e)=g$, so fixing the identity element already forces $g=e$. In fact every stabilizer subgroup is trivial, since $gx=x$ forces $g=e$; this stronger property makes the left regular action a regular group action.^[ref]

Compare two group actions of the same cyclic group $C_4$. Its left regular action on four labels distinguishes all four group elements. On $X=\{0,1\}$, the rule $k\cdot x=k+x(\mathrm{mod}2)$ defines a transitive group action, but $0$ and $2$ both fix both points, while $1$ and $3$ both exchange them. Its group kernel is $\{0,2\}$, so it is not a faithful group action. Its image is a two-element subgroup of $S_2$, isomorphic to the quotient group $C_4/\{0,2\}$ rather than to $C_4$.

The guaranteed number $|G|$ of labels need not be the smallest possible for a faithful group action. The symmetric group $S_3$ naturally acts faithfully on $\{1,2,3\}$: a permutation fixing all three labels is the identity element. Its left regular action uses six labels instead. Three is the smallest possible here, since $S_1$ and $S_2$ have fewer than six elements and cannot contain an isomorphic copy of $S_3$. Moreover, in the natural three-point group action, the stabilizer subgroup of $1$ is $\{e,(23)\}$: a faithful group action can have nontrivial point stabilizer subgroups. Cayley's construction supplies a universal model, while a smaller model can reveal particular symmetries.

The proof is short. Each $L_g$ is a bijection with inverse $L_{g^{-1}}$. Associativity givesso $\lambda$ is a group homomorphism. If $L_g=L_h$, evaluating both at the identity element gives $g=L_g(e)=L_h(e)=h$. Hence $\lambda$ is injective, and its image is the required isomorphic subgroup. No finiteness assumption entered the argument.

The historical connection is unusually close to the notation. The opening paragraph of Arthur Cayley's 1854 paper already treats permutations as one example of operations on a collection of quantities and explicitly performs the rightmost operation first in a product. It also allows noncommutativity while insisting on associativity. The modern left regular action shows how that viewpoint reaches every abstract group, by choosing the group itself as the set to be acted on.^[ref]

A Cayley table records a binary operation on a finite set: the cell at row $a$, column $b$ contains $a\ast b$. Every cell must contain an element of the specified set. With the same ordering of row and column labels, the table is symmetric across its main diagonal exactly when the binary operation is commutative. Checking the associative property instead requires comparing $(a\ast b)\ast c$ with $a\ast (b\ast c)$ for every triple; symmetry alone cannot establish it.^[ref]

Both constructions are stabilizer subgroups for the conjugation action. The centralizer $C_G(x)$ fixes an individual element $x$; the normalizer $N_G(H)$ preserves a subgroup $H$ as a whole, while possibly moving its elements. The centralizer of a subset $C_G(H)$ fixes every member of $H$ individually, soFor a cyclic subgroup $H=⟨r⟩$, commuting with the group generator $r$ is equivalent to commuting with all its powers, giving $C_G(H)=C_G(r)$.

In the symmetric group $S_3$, let $r=(123)$ and $H=⟨r⟩=\{e,(123),(132)\}$. ThenEvery transposition sends $r$ to $r^{-1}$ under conjugation. In particular, for $t=(12)$,Thus $t$ belongs to the normalizer of $H$ but not to the centralizer of $r$. Preserving the collection of powers allows the two possible directions of its group generator to be exchanged.

This distinction lets one ask separately which operations leave a chosen operation unchanged and which operations preserve the subgroup it generates. The normalizer also supplies a group homomorphism to the automorphism group of $H$, by $g\mapsto (h\mapsto gh{g}^{-1})$; its group kernel is $C_G(H)$. In this example the group homomorphism has two possible outputs, fixing $r$ or replacing it by $r^{-1}$. The quotient group $N_{S_3}(H)/C_{S_3}(H)$ is therefore a cyclic group of group order two. Milne develops the normalizer as a stabilizer subgroup in Group Theory (v4.01) by James S. Milne.

For a finite group $G$, choose one representative $x_i$ from each conjugacy class with more than one element. The class equation iswhere $Z(G)$ is the center of a group and $C_G(x_i)$ is the centralizer of $x_i$. Each central element contributes its own one-element conjugacy class. Every remaining conjugacy class contributes its size, given by the orbit-stabilizer theorem. The sum counts each element exactly once. See McKernan, Group actions and automorphisms, lemma 23.6.

For conjugacy classes of the symmetric group on three letters, take $x_1=(12)$ and $x_2=(123)$. Their centralizers have sizes two and three, and the center of a group has size one, givingFor an abelian group, every centralizer is the whole group, every conjugacy class has size one, and the sum over noncentral representatives is empty.

The power of this counting identity is that the sizes of conjugacy classes constrain group structure. For a nontrivial finite p-group with group order $p^n$, every noncentral conjugacy class has size a positive power of $p$, by Lagrange's theorem. The class equation therefore gives $|Z(G)|\equiv 0(\mathrm{mod}p)$. Since the identity element lies in the center of a group, the center of a group must have at least $p$ elements. This obtains an element commuting with the entire group without constructing one. See McKernan, lemma 23.7.

Historically, these ideas belong to the nineteenth-century development of permutation groups and normal subgroups. Milne's historical opening records Galois's introduction of normal subgroups in 1832 and Jordan's systematic treatment of permutation groups in 1870. This is context for the conjugation action and the class equation, rather than a claim that either author first stated the modern formula.^[ref]

For a subgroup $H$ of a group $G$ and $g\in G$, the conjugate subgroup is the conjugate subsetIt is a subgroup because conjugation is a group automorphism. Restricting that group automorphism gives a group isomorphism from $H$ to $gH{g}^{-1}$, so conjugate subgroups have the same group order. This describes the same operations after moving their reference configuration. Conjugacy of subgroups is an equivalence relation. ^[ref]

For an equilateral triangle with vertices of a polygon labeled $1,2,3$ counterclockwise, its symmetry group is the symmetric group $S_3$, also the dihedral group of group order six. Let $r=(123)$ be the counterclockwise rotation through $120$ degrees, and let $s=(12)$ be the reflection fixing vertex of a polygon $3$. Products apply the rightmost permutation first. The cyclic subgroup $K=⟨s⟩=\{e,(12)\}$ contains the identity element and one reflection. Relabeling a transposition by conjugation givesThus the conjugation action moves $K$ among the three subgroups $⟨(12)⟩$, $⟨(23)⟩$, and $⟨(13)⟩$. Geometrically, the rotation moves the axis of the reflection to the next axis. Each reflection fixes a different vertex of a polygon, although group isomorphisms relate all three subgroups.

The normalizer tells when this change of reference leaves the subgroup unchanged. For any group $G$ and subgroup $H\le G$, there is a bijectionTwo choices $g,k$ give the same conjugate subgroup exactly when $k^{-1}g\in N_G(H)$. Consequently the index of a subgroup $[G:N_G(H)]$ counts the distinct conjugate subgroups, even for an infinite group. For a finite group the count is $|G|/|N_G(H)|$. These are cosets; they need not form a quotient group, since $N_G(H)$ need not be a normal subgroup of $G$. In the triangle example there are $6/2=3$ conjugate subgroups, while the cyclic subgroup $⟨r⟩$ has only one. ^[ref]

A group acts on the left cosets of a subgroup by left multiplication; its kernel is the core of that subgroup.

Let $H\le G$ be a subgroup. The notation $G/H$ here means the set of left cosets $\{aH:a\in G\}$, even when $H$ is not a normal subgroup. DefineThis gives a group action without needing a quotient group structure on $G/H$.^[ref]

The representative check is essential: if $aH=bH$, then $a=bh$ for some $h\in H$, and $(ga)H=(gbh)H=(gb)H$. Thus equal input cosets give equal output cosets. The identity element fixes every coset, and associativity gives $g_1\cdot (g_2\cdot aH)=(g_1{g}_2)\cdot aH$. Each induced map is a permutation, with inverse induced by $g^{-1}$. This is a transitive group action because $b{a}^{-1}$ sends $aH$ to $bH$.

The stabilizer subgroup of $H$ is $H$ itself. At a general coset, the stabilizer subgroup and the group kernel areConsequently the coset action is a faithful group action exactly when the subgroup core is trivial.^[ref]

Indeed, $g\cdot (aH)=aH$ exactly when $a^{-1}ga\in H$, or $g\in aH{a}^{-1}$. Intersecting these membership conditions over all cosets gives the group kernel. Conjugation permutes the subgroups in this intersection, making it a normal subgroup of $G$ contained in $H$. Conversely, any normal subgroup of $G$ contained in $H$ lies in every $aH{a}^{-1}$. This explains why the subgroup core is the largest such normal subgroup: it is exactly the portion of $H$ invisible at every label, rather than merely at the label $H$.

For a normal subgroup $H$, every $aH{a}^{-1}$ equals $H$, so $\ker {\rho }_H=H$. For $H=\{e\}$, the construction is the regular group action behind Cayley's theorem. Between these extremes, a nontrivial subgroup with trivial subgroup core can give a faithful group action on fewer labels. If the index of a subgroup is finite, $[G:H]=n$, the first isomorphism theorem givesThus a subgroup of small index of a subgroup exposes a small permutation group associated with $G$. The two examples below and in nonfaithful coset action of the symmetric group on three letters show exactly which information can survive.

A cyclic group is a group that can be generated by repeating a single group element $g$ and its inverse element:For an additive group, the same expression is written $\{kg:k\in \mathbb{Z}\}$: $kg$ means repeated addition, including subtraction when $k$ is negative. A finite cyclic group has a repeating cycle, while an infinite cyclic group is isomorphic to the additive integers. The notation $C_n$ refers to a finite cyclic group of order $n$.

This definition compresses a whole composition rule into one repeatable operation. Once $g$ is chosen, $g^a{g}^b=g^{a+b}$, so working with group elements becomes working with integer exponents. In the finite case those exponents are reduced using modular arithmetic. It also proves that every cyclic group is an abelian group, because $a+b=b+a$. A finite cyclic group of order $n$ has exactly $d$ solutions of $x^d=e$ for every positive integer divisor $d$ of $n$: in exponents, $n\mid dk$ has exactly $d$ solutions modulo $n$. The standard finite and infinite models are described in Group Theory (v4.01) by James S. Milne (section 1.16).

Every endomorphism of the additive finite cyclic group $C_n=\mathbb{Z}/n\mathbb{Z}$, for a positive integer $n$, has the form $f_a(k)=ak(\mathrm{mod}n)$, determined by $a=f_a(1)$. This is the self-map case of homomorphism from a finite cyclic group: repeated addition forces the value at every element. Conversely, each choice of $a$ respects addition and gives a well-defined map on congruence classes.

Let $d=\gcd (n,a)$, the greatest common divisor. The image of a group homomorphism is the cyclic subgroup generated by $a$, with size $n/d$ by order of an element of a finite cyclic group. The kernel of a group homomorphism isIndeed, $ak=0(\mathrm{mod}n)$ is equivalent to $n/d$ dividing $k$, after cancelling $d$ and using coprimality. Consequently $f_a$ is a group automorphism exactly when $a$ and $n$ are coprime. In that case a modular inverse $b$ of $a$ gives $f_a^{-1}=f_b$. This connects reversible self-maps directly to the generators of a finite cyclic group.^[ref]

A finite abelian group combines a finite set of reversible operations with commutativity. All finite cyclic groups qualify, but direct products of groups supply examples that need more than one generator of a group. The Fundamental theorem of finitely generated abelian groups says that every finite abelian group is a direct product of groups of finite cyclic groups; the finite case has no infinite cyclic factors.^[ref]

These factors distinguish different ways to have the same total number of states. A single four-position dial gives $C_4$, while two independent binary switches give $C_2\times C_2$. Both have four states and commutative composition, yet only the dial permits a single operation whose repetition visits all four states. The following Cayley tables make that distinction visible without relying on the labels alone.

The free monoid ${\Sigma }^{\ast }$ on an alphabet $\Sigma$ consists of all finite words over an alphabet, including the empty word $\epsilon$, with concatenation as its binary operation. Distinct strings represent distinct elements: there are no imposed equalities between different sequences of letters. The empty word is its identity element, and concatenation has the associative property.^[ref]

For $\Sigma =\{a,b\}$, the elements begin $\epsilon ,a,b,aa,ab,ba,bb,\dots$. Both $(ab)a$ and $a(ba)$ give the same word $aba$, but $ab\ne ba$, so this monoid is not commutative. The length of a word over an alphabet obeys $\ell (uv)=\ell (u)+\ell (v)$; it is a monoid homomorphism to the nonnegative integers with addition. A nonempty word has no inverse element, since joining it to any other word cannot decrease its length to zero.

The defining benefit of being free is that letters can stand for arbitrary elements of another monoid. Choose a value $h(a)$ for each letter $a\in \Sigma$; then $h(a_1\cdots a_n)=h(a_1)\ast \cdots \ast h(a_n)$, with $h(\epsilon )=e$, is the unique monoid homomorphism extending those choices. Associativity and the identity element make this evaluation consistent for every finite word. For example, interpreting the letters as self-maps evaluates a word as a composed transformation; interpreting them as nonnegative counts evaluates it by addition.

A group packages reversible operations together with a rule for composing them. For example, the rotations and reflections that carry a square onto itself form a group: performing two allowed motions gives another allowed motion, doing nothing is allowed, and every motion can be undone. The elements are the motions themselves, rather than the vertices they move. Labels in the figure track those vertices; they are markers, rather than extra decorations that the motions must preserve.

Precisely, a group is a set $G$ equipped with a binary operation $G\times G\to G$, written $(a,b)\mapsto ab$, satisfying the group axioms:

The group operation need not be ordinary numerical multiplication. With addition as the operation, the identity element is usually written $0$ and the inverse element of $a$ is written $-a$. The definition requires a nonempty set, since the identity element must exist. See McKernan's MIT lecture 1, Definition 1.1 for the axioms.

Why insist on associativity? It makes a string of operations unambiguous without specifying parentheses. For function composition, both $(fg)h$ and $f(gh)$ send $x$ to $f(g(h(x)))$. This fixes the order of execution while allowing regrouping. Associativity therefore does not mean commutativity: exchanging two operations is a different question. Ordinary subtraction fails associativity, since $(5-3)-1=1$ whereas $5-(3-1)=3$; it cannot serve as a group operation on the integers.

An inverse element makes reversal exact and equations solvable. In any group, $ax=b$ has the unique solution $x=a^{-1}b$, since $a^{-1}(ax)=(a^{-1}a)x=x$. Similarly, $xa=b$ has solution $x=b{a}^{-1}$. The side matters in a non-abelian group. Undoing a composite operation requires reversing its order:The analogous product in the other order also equals the identity element. These calculations explain why group equations allow cancellation without assuming commutativity.

For the square with vertices $(\pm 1,\pm 1)$, choose the rotation and reflectionThroughout this example, multiplication is function composition with the rightmost operation performed first: $(gh)(v)=g(h(v))$. Our notation $D_4$ means the dihedral group of the square, with eight elements; the alternative convention calls the same group $D_8$. Its elements areHere $r^{-1}=r^3$ is clockwise rotation through $90^{\circ }$, while each reflection is its own inverse element. There are exactly eight motions: the image of one vertex has four choices, and the image of an adjacent vertex has two choices; together these determine the motion. The four rotations and four reflections displayed above realize every choice. The vertex permutations give a concrete group action. The general polygon construction is discussed in Group Theory (v4.01) by James S. Milne (section 1.17).

Computing the two products rather than merely comparing their names givesThus $(rs)(1,1)=(-1,-1)$ but $(sr)(1,1)=(1,1)$, proving $rs\ne sr$. In fact $rs$ is reflection in the line $y=-x$, whereas $sr$ is reflection in $y=x$. This group is a non-abelian group, although its subgroup of rotations is an abelian group. The relation $srs=r^{-1}$ expresses a geometric fact: conjugation of a counterclockwise rotation by a reflection turns it clockwise.

A contrasting example is the cyclic group $\mathbb{Z}/4\mathbb{Z}$ under addition in modular arithmetic. Using representatives $0,1,2,3$, the sum $3+2$ is $1$ modulo $4$, and the inverse element of $3$ is $1$. Repeated addition of $1$ runs through $0,1,2,3,0$. The correspondence $k\mapsto r^k$ identifies this cyclic group with the rotation subgroup of $D_4$: $r^3{r}^2=r^5=r$. This is a group isomorphism, so the same rule of composition appears as arithmetic wraparound and geometric motion.

By comparison, $(\mathbb{Z},+)$ is an infinite cyclic group. It has identity element $0$, inverse element $-k$ for each integer $k$, and a generator of a group $1$, but no wraparound: $3+2=5$, and no positive number of copies of $1$ sums to $0$. Both examples are abelian groups. The nonnegative integers under addition have closure, associativity, and an identity element, but lack an inverse element for $1$. Reversibility is precisely what fails.

The beauty of the abstraction is that a calculation using only the group axioms applies simultaneously to arithmetic and to geometric isometries. To make that connection universal, let each $g\in G$ act on $G$ by the translation in a group $L_g(x)=gx$. Each $L_g$ is a bijection, with inverse $L_{g^{-1}}$, and $L_g{L}_h=L_{gh}$. Distinct elements give distinct maps because $L_g(e)=g$. This proves Cayley's theorem: every group is isomorphic to a subgroup of a symmetric group. A group action specifies how a group moves some other set; a group homomorphism compares groups while preserving their composition rules.

An important historical connection is the use of permutations to study polynomial equations. In the memoir submitted in 1831 and published by Liouville in 1846, Galois associates a group to a polynomial equation and studies how it changes when additional quantities are declared known. Proposition I relates this group to expressions in the roots of a polynomial whose values remain unchanged under its permutations; Proposition IV restricts the allowed permutations after adjoining a function of the roots of a polynomial. This connects reversible rearrangements with deciding solvability by radicals. See Galois's original memoir, pp. 417–433, especially Propositions I and IV; Liouville's introduction, p. 382.

A group automorphism is a bijective group homomorphism $\alpha :G\to G$. It preserves the group operation, the identity element, and inverse elements; its inverse is another group homomorphism. Thus it is a group isomorphism with identical source and target. These maps form the automorphism group $\mathrm{Aut}(G)$ under function composition.^[ref]

For the additive finite cyclic group $C_6$, take $f_5(k)=5k(\mathrm{mod}6)$. Its output list on $0,1,2,3,4,5$ is $0,5,4,3,2,1$, a permutation of all six elements. It preserves addition because $5(a+b)=5a+5b(\mathrm{mod}6)$, and applying it twice gives $25k=k(\mathrm{mod}6)$. Hence it is a group automorphism and its own inverse function. Compare the collapsing $f_2$ in the figure under endomorphism. The general criterion under endomorphism of a finite cyclic group says exactly when such a map is reversible: its multiplier must be coprime to the size of the finite cyclic group.

A permutation need not be a group automorphism, even if it fixes the identity element. Let $\pi$ exchange $1$ and $2$ in $C_6$ and fix $0,3,4,5$. It is a bijection, and $\pi (0)=0$, butThus it fails to preserve addition and is not even a group homomorphism. Reversibility alone does not preserve the group operation.

A group automorphism carries each cyclic subgroup onto the cyclic subgroup generated by the image of its defining element, preserving the order of a group element. In particular, it carries a generator of a group to another generator of a group. For $C_6$, the existing calculation of generators of a finite cyclic group leaves only the images $1$ and $5$ for $1$, so its automorphism group consists exactly of the identity function and $f_5$. This turns the problem of compatible relabeling into a much smaller choice than enumerating arbitrary permutations.

A group isomorphism is a bijective group homomorphism $\phi :G\to H$. It matches the entire group operation: $\phi (ab)=\phi (a)\phi (b)$, with each product taken in its own group. Its inverse is automatically a group homomorphism. A group automorphism is the special case $H=G$.^[ref]

The target is part of the claim. The doubling example under isomorphism is a group isomorphism $d:\mathbb{Z}\to 2\mathbb{Z}$, between the additive integers and the additive even integers. Exactly the same rule $d(k)=2k$, declared as $d:\mathbb{Z}\to \mathbb{Z}$, is an injective endomorphism that misses $1$ and every other odd integer, so is not a group automorphism. Changing the declared target changes surjectivity, even though it does not change the formula or its compatibility with addition.

An isomorphism lets a calculation in one model solve the same problem in another. The example under group identifies addition modulo four with function composition of rotations of a square via $k\mapsto r^k$. Through this group isomorphism, adding exponents corresponds exactly to composing motions. More generally, a group isomorphism preserves order of a group element, carries subgroups to subgroups, and carries a generating set of a group to a generating set of a group. Having equally many elements is insufficient: the permutation in group automorphism already fails compatibility on two inputs.

Why does the inverse preserve the group operation? Write $u=\phi (a)$ and $v=\phi (b)$, possible because of surjectivity. ThenThis proves the inverse condition directly. Analogous calculations work for each operation of a ring and for addition and scalar multiplication in a vector space. The argument relies on preserving those operations; it supplies no general continuity claim for the inverse of a continuous bijection between topological spaces.

A monoid is a semigroup with a two-sided identity element $e$: $e\ast a=a\ast e=a$ for every element $a$. Associativity handles the grouping of a nonempty sequence; the identity element also supplies the result for an empty sequence. This makes the same rules apply to combining zero, one, or many contributions. ^[ref].

The nonnegative integers $\{0,1,2,\dots \}$ with addition form a commutative monoid, with identity element zero. They do not form a group: a positive integer has no additive inverse in that set. All words over an alphabet with concatenation form a free monoid, whose identity element is the empty word. The full transformation monoid consists of all self-maps of a set, with function composition and the identity function.

These examples explain why monoids are useful: accumulation needs an operation and a starting value, but it need not be reversible. Adding counts, joining strings, and composing transformations all share the same grouping and identity laws. Requiring every element to have an inverse element gives a group, a stronger structure suitable for reversible operations.

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.