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In a finite monomial group, the displayed intersection is a normal subgroup. Every irreducible character is induced from a linear character of a subgroup containing $K$. On restriction to $K$, all conjugate inducing lines remain invariant, so $K^{\prime }$ lies in every irreducible character kernel. The intersection of those kernels is trivial, proving that $K$ is Abelian.

Use the coordinate conventionIn this convention the Heisenberg group over a prime field has the productThe subset $H=\{A(t,b,t):t,b\in {\mathbb{F}}_p\}$ is an Abelian subgroup of order $p^2$: the middle coordinate in a product is $b+b^{\prime }+t{t}^{\prime }$, which is symmetric in the factors. Also $A(t,b,t{)}^{-1}=A(-t,t^2-b,-t)$ lies in $H$. The map $A(a,b,c)\mapsto a-c$ is a surjective group homomorphism to the additive group of ${\mathbb{F}}_p$, with kernel of a group homomorphism $H$. Hence $H$ is a normal subgroup of subgroup index $p$, and its left cosets are exactly the level sets of $a-c$.

For odd $p$, an explicit group isomorphism from $H$ to the additive group ${\mathbb{F}}_p^2$ isThe term $t{t}^{\prime }$ in matrix multiplication cancels the cross term in $(t+t^{\prime }{)}^2/2$. Thus $H$ is an elementary abelian group, isomorphic to $C_p\times C_p$.

In the Heisenberg group over a commutative ring with $R=\mathbb{Z}/4\mathbb{Z}$, the subset $H=\{A(t,y,t):t,y\in R\}$ is an Abelian subgroup of group order 16. Its product has central coordinate $y+y^{\prime }+t{t}^{\prime }$, symmetric in the two factors. Its inverse element is $A(-t,t^2-y,-t)$. The map $A(x,y,z)\mapsto x-z$ is a surjective group homomorphism to the additive group of $R$, with kernel of a group homomorphism $H$. Thus $H$ is a normal subgroup and has four cosets in the 64-element group.

Unlike the odd-prime case, division by two is unavailable. In fact take $u=A(1,0,1)$ and $v=A(2,0,2)$. Then $u^n=A(n,(\genfrac{}{}{0ex}{}{n}{2}),n)$, where the binomial coefficient is computed as an integer before reduction modulo four. Hence $u^4=A(0,2,0)$ and $u$ has element order eight. The element $v$ has element order two and is not a power of $u$: the powers with outer coordinates two have central coordinates one or three. Their generated subgroups intersect trivially, so the 16-element Abelian group is the direct product of groups $C_8\times C_2$.

An abelian quotient is a quotient group $G/N$ that is an abelian group. It exists only for a normal subgroup $N$, and it is abelian exactly when $N$ contains the commutator subgroup. Thus every abelian quotient factors through the abelianization $G/[G,G]$.

The Sylow theorems give either one or eight Sylow subgroups of order seven. In the first case that subgroup is normal. In the second, their six nonidentity elements are pairwise disjoint, accounting for 48 elements. Only eight elements remain, including the identity, and every Sylow subgroup of order eight lies entirely among them. Thus there is exactly one, and it is a normal subgroup.

For a finite simple group with a subgroup of index $n\ge 3$, the coset action is nontrivial and its kernel is a normal subgroup, hence trivial. Its sign homomorphism must also be trivial: otherwise simplicity would give an isomorphism to $C_2$, which cannot act transitively on at least three points. Thus the image lies in the alternating group. In particular Lagrange's theorem gives $|G|\mid n!/2$. This can exclude simplicity with very little counting: a group of order $112$ has a subgroup of index seven by Sylow theorems, but $112$ does not divide $7!/2$.

The center of a group consists of the elements that commute with every element of that group:where $C_G(g)$ is the centralizer of $g$. These are operations whose position in any product can be changed freely. The center is an abelian group and a characteristic subgroup, hence a normal subgroup. A group is an abelian group precisely when $Z(G)=G$.

For the symmetric group $S_3$, the conjugacy classes areAn element belongs to the center precisely when its conjugacy class has one element, so $Z(S_3)=\{e\}$. Yet commutator subgroup of the symmetric group on three letters gives $S_3^{\prime }=A_3$, with three elements. Thus belonging to the commutator subgroup does not mean commuting with everything. The alternating group $A_3$ is an abelian group as a subgroup, but its nonidentity elements do not commute with the transpositions in the surrounding $S_3$.

For the dihedral group $D_8$ of group order eight in abelianization of the square symmetry group, the center is $\{e,r^2\}$. A rotation $r^i$ commutes with $s$ precisely when $r^i=r^{-i}$, which requires $i$ even. No reflection $r^{i}s$ commutes with $r$, since $(r^{i}s)r=r^{i-1}s$ whereas $r(r^{i}s)=r^{i+1}s$. ConsequentlyThe half-turn can be performed before or after any square symmetry with the same result. Its appearance both here and in the commutator subgroup is a feature of this example, rather than a general equality. See Conrad, Dihedral groups I, theorem 3.4 and example 3.5.

The center and the commutator subgroup answer different useful questions: which operations already commute with everything, and which operations must disappear for the entire group to commute. In an abelian group these answers are $Z(G)=G$ and $G^{\prime }=\{e\}$. Another way to use the center is through conjugation: the group homomorphism $g\mapsto (h\mapsto gh{g}^{-1})$ has group kernel $Z(G)$, so the first isomorphism theorem gives $G/Z(G)\cong \mathrm{Inn}(G)$, the inner automorphism group. The quotient group by the center thus records the distinct conjugation operations; it need not be an abelian quotient.

A nontrivial finite p-group always has a nontrivial center. In the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. The identity element ensures $Z(G)$ is nonempty, hence it has at least $p$ elements. See McKernan, MIT 18.703, lemmas 23.6–23.7 and definition-lemma 23.10.

A nontrivial normal subgroup $N$ of a finite p-group meets the center of a group nontrivially. The conjugation action of $G$ on $N$ has singleton orbits exactly at $N\cap Z(G)$. The orbit-stabilizer theorem makes every other orbit size divisible by $p$, so $|N\cap Z(G)|\equiv |N|\equiv 0(\mathrm{mod}p)$. Since the intersection contains the identity, its size is at least $p$.

In the conjugacy class order $1,(123),(12)(34),(12345),(13452)$, the five complex irreducible characters of the alternating group $A_5$ areThe class sizes are $1,20,15,12,12$. Restricting the degree-four and degree-five irreducible characters of $S_5$ gives rows of squared character inner product one. The degree-six restriction has squared norm two and is orthogonal to those rows and the trivial row, so it splits into the two remaining irreducibles. The sum of squared degrees forces their degrees to be three. Orthogonality forces their last two values to have sum one and product minus one, producing the displayed roots. Since a normal subgroup is a union of classes and its order divides sixty, these class sizes also prove the Simplicity of the alternating group A5.

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]

The commutator subgroup of a group $G$ is the subgroup generated by all its group commutators:Here and throughout, $[x,y]=x^{-1}{y}^{-1}xy$. A group commutator is the identity element precisely when $x$ and $y$ commute. The generated subgroup includes finite products of group commutators and their inverse elements; an element of $G^{\prime }$ need not itself be a single group commutator. Thus $G^{\prime }=\{e\}$ precisely when $G$ is an abelian group.

The useful idea is to record exactly what must disappear to make multiplication commutative. The identity $xy=yx[x,y]$ shows how a group commutator corrects a change in multiplication order. Setting every such correction equal to $e$ produces the abelianization $G/G^{\prime }$, while preserving every piece of information that a group homomorphism to an abelian group can detect. This connects a local comparison of two operations to a canonical quotient group of the whole group.

The commutator subgroup is a characteristic subgroup, hence a normal subgroup: a group automorphism $\alpha$ sends $[x,y]$ to $[\alpha (x),\alpha (y)]$. For any normal subgroup $N$,Indeed, the group commutator of $xN$ and $yN$ is $[x,y]N$. This makes $G^{\prime }$ the smallest normal subgroup whose quotient group is an abelian group. See Conrad, Subgroup series I, section 4: commutators and abelian quotients.

A group complement to a normal subgroup $N◃G$ is a subgroup $H$ with $G=NH$ and $N\cap H=1$. Projection identifies $H$ with $G/N$, and conjugation gives $G=N⋊H$. This is a subgroup notion, not the set-theoretic complement of $N$.

The alternating conjugacy class splitting criterion gives seven conjugacy classes in $A_6$. Their cycle types and sizes are $1^6:1$, $2^2{1}^2:45$, $31^3:40$, $3^2:40$, $42:90$, and two $51$ classes of size 72. Representatives of the split classes are a 5-cycle and its square. Their sizes sum to 360. A normal subgroup must be a union of these classes containing the identity, so these explicit sizes can prove simplicity using Lagrange's theorem.

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]

The core of a subgroup $H\le G$ isIt is the largest normal subgroup of $G$ contained in $H$. If $H$ has finite index $k$, the action on $G/H$ gives $[G:{\mathrm{Core}}_G(H)]\le k!$.

For a subgroup $H\le G$ of a group $G$, a left coset is $gH=\{gh:h\in H\}$; a right coset is $Hg=\{hg:h\in H\}$. The name refers to the side on which the representative $g$ multiplies $H$. Think of a coset as a translated copy of a subgroup: it remembers the relative differences allowed by $H$, while moving the whole collection to another location in the group.

Two left cosets of the same subgroup are either equal or disjoint. Thus distinct left cosets form a partition of a set covering $G$; the same statement holds separately for right cosets. The useful membership test isThis also defines an equivalence relation: changing a representative by right multiplication with a member of $H$ leaves its left coset unchanged. These facts hold even for infinite groups. See MIT's coset notes, definitions 3.6 and 3.9.

For an explicit additive example, take the cyclic group $G=\mathbb{Z}/6\mathbb{Z}$, with modular arithmetic, and the subgroup $H=\{0,3\}$. Adding each representative to both members gives exactlyRepresentatives $3,4,5$ repeat these three cosets, respectively. For example, $4+H=\{4,1\}=1+H$, since $4-1=3\in H$. Every residue appears once in the displayed partition of a set. Each coset has two members, so the group order of $H$ is $2$ and its index of a subgroup is $3$. Left and right cosets coincide here because $G$ is an abelian group.

A coset need not be a subgroup. In this example, $\{1,4\}$ lacks the identity element $0$, and $1+1=2$ leaves it. In fact, the only left or right coset of $H$ that is a subgroup is $H$ itself: a subgroup must contain the identity element, and the coset containing that element is $H$. Equal sizes come instead from the bijection $h\mapsto gh$, whose inverse is $x\mapsto g^{-1}x$; for right cosets, use $h\mapsto hg$ and $x\mapsto x{g}^{-1}$.

The nonabelian group $S_3$, the symmetric group on three letters, separates the two sides. Use function composition with the rightmost permutation applied first, and let $H=\{e,(12)\}$. Then $(13)(12)=(123)$ but $(12)(13)=(132)$, giving the differing pairThe complete partition of a set into left cosets consists of $H$, $\{(13),(123)\}$ and $\{(23),(132)\}$; the complete partition of a set into right cosets consists of $H$, $\{(13),(132)\}$ and $\{(23),(123)\}$. Each lists all six permutations exactly once. Both have three parts of size two, although their parts differ.

Left cosets support a group action by $a\cdot (gH)=(ag)H$ for every subgroup, as in the coset action. Multiplication of representatives defines a quotient group operation $(aH)(bH)=(ab)H$ precisely when $H$ is a normal subgroup. Otherwise the answer can depend on the representatives: in this $S_3$ example, $e$ and $(12)$ represent the same coset $H$, yet $e(13)H=(13)H$ and $(12)(13)H=(23)H$ differ. In the additive example, $H$ is a normal subgroup, making $(1+H)+(2+H)=3+H=H$ valid. See MIT's quotient-group notes, theorem 9.1.

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.

If $N$ is a normal subgroup, changing representatives to $gn$ and $hm$, where $n,m\in N$, givesThe chosen representatives therefore give the same coset. Conversely, suppose the rule defines an operation on left cosets. Since $nN=N$ for $n\in N$, multiplying by $gN$ gives $ngN=gN$, hence $g^{-1}ng\in N$. Applying this inclusion for every $g$ and $g^{-1}$ proves $g^{-1}Ng=N$, so $N$ is a normal subgroup. This makes normality exactly the condition needed to carry group operation to these classes.

Write $M(a,b,c)=\left(\begin{array}{ccc}1& a& b\\ 0& 1& c\\ 0& 0& 1\end{array}\right)$ with integer coordinates. Multiplication in this parametrization is $M(a,b,c)M(A,B,C)=M(a+A,b+B+aC,c+C)$. The subgroup $H=\{M(0,0,t):t\in \mathbb{Z}\}$ has right cosets obtained by multiplying on the left by its elements, so its right cosets are exactly the displayed sets with fixed $(a,b)$. In particular, right cosets are indexed by ${\mathbb{Z}}^2$. Left cosets differ: $M(a,b,c)H$ fixes $a$ and $b-ac$, since right multiplication changes $b$ by $at$ and $c$ by $t$. This also exhibits why $H$ is not a normal subgroup of the Integer Heisenberg group.

If $V$ is a value in a lattice-ordered group of $t>1$, every convex lattice subgroup strictly containing $V$ contains $t$. Thus the convex lattice subgroup generated by $V$ and $t$ is the unique least subgroup properly containing $V$. It is called the cover of $V$. There is no intermediate convex lattice subgroup. A normal-valued lattice-ordered group requires $V$ to be a normal subgroup of this cover, not necessarily of the whole group. Values are prime convex lattice subgroups: maximality omitting $t$ makes them meet-irreducible in the lattice of convex lattice subgroups, so their ordered right cosets form a chain.