Finite group

A finite group is a group whose underlying set has finitely many elements. Its order of a finite group is the number of elements, counting the identity element. The group operation still has to satisfy all the group axioms: merely having a finite list of states does not make a group.

For example, the six congruence classes modulo six form a finite cyclic group under addition, whereas the permutations of three objects form the symmetric group $S_3$, also with six elements. The first is an abelian group; the second is a non-abelian group. Thus the order of a finite group is a useful first measurement, but the composition rule contains information that the count alone loses.

Finiteness makes group structure accessible to exact calculation. A Cayley table for a finite group of order $n$ has $n^2$ entries, and every group element eventually returns to the identity element when repeated. Lagrange's theorem strengthens this observation: the order of a group element, being the size of its cyclic subgroup, divides $n$. The cyclic group of order four and Klein four-group give a small example of how that constraint can distinguish different composition rules.

The p-prime core is the largest normal subgroup whose order is coprime to $p$. Products of normal subgroups of order coprime to $p$ again have order coprime to $p$, so it exists and is characteristic.

The p-core is the largest normal p-subgroup of a finite group. The product of two normal p-subgroups is again a normal p-subgroup, so this subgroup exists and is characteristic.

Put $\overline{G}=G/O_{p^{\prime }}(G)$ using the p-prime core of a finite group. A finite group is p-constrained if $C_{\overline{G}}(O_p(\overline{G}))\le O_p(\overline{G})$. If $O_{p^{\prime }}(G)=1$, the p-core of a finite group has its centralizer contained in itself. Then block idempotents centralize a normal p-subgroup places every block idempotent in $k{O}_p(G)$, a local ring, so $kG$ has only the principal block.

For a set $\pi$ of primes, a pi-group is a finite group whose order has no prime divisors outside $\pi$. The trivial group qualifies for every $\pi$.

A pi-element is an element whose order has no prime divisors outside a specified prime set $\pi$. The identity is a pi-element for every prime set. Conjugation preserves the property because it preserves element order.

In finite-group character theory, an elementary group is a group that is p-elementary group for some prime $p$. Such groups are solvable and provide the subgroup tests in Brauer's characterization of characters. This does not mean elementary abelian group.

Write an elementary group as $P\times C$ with $P$ a p-group and $C$ cyclic of order prime to $p$. Split $C$ into its cyclic pi-part and pi-complementary part. Attach $P$ to whichever part contains $p$, producing a direct product of groups of a pi-group and a complementary-prime group.

A p-elementary group is a direct product of groups of a p-group and a cyclic group whose order is coprime to $p$. Either factor may be trivial. The coprime cyclic factor is central, a stronger requirement than merely having a cyclic subgroup with p-group quotient.

A finite group is ambivalent when every element is conjugate to its inverse. Since every complex character satisfies $\chi (g^{-1})=\overline{\chi (g)}$, an ambivalent group has real values for every irreducible character. Conversely, if every irreducible value is real, irreducible characters separate conjugacy classes and force every $g$ to be conjugate to $g^{-1}$.

A finite group is monomial if every irreducible character is a monomial character. Finite metabelian groups have this property.

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.

Every finite solvable group is isomorphic to a subgroup of a finite monomial group. This is an embedding assertion and does not assert that every solvable group is monomial, nor that the embedding is normal.

Every finite monomial group is a solvable group. Thus expressing every irreducible character as induction from a linear character imposes a group-theoretic solvability condition.

A Frobenius group has a nontrivial proper subgroup $H$ satisfying $H\cap H^g=\{1\}$ for every $g\notin H$, with $H^g=g^{-1}Hg$. Such $H$ is a Frobenius complement. The Frobenius kernel theorem turns the identity together with elements outside all conjugates of $H$ into a normal subgroup $N$, giving $G=N⋊H$.

The Frobenius kernel consists of the identity and the elements lying in no conjugate of a Frobenius complement. The Frobenius kernel theorem proves that this conjugacy-invariant set is a normal subgroup, rather than assuming closure as part of its definition.

For $1<N◃G$ proper, the displayed condition is equivalent to $N$ being a Frobenius kernel. For a normal subgroup the induction norm is $|I_G(\phi ):N|$. Trivial inertia for each nonprincipal character and Brauer's permutation lemma imply that every element outside $N$ has trivial centralizer in $N$. This forces $N$ to be a normal Hall subgroup; a complement is then fixed-point-free on $N$.

Let $N$ be a finite group and $\alpha$ an automorphism whose only fixed element is the identity. If $\alpha (a{)}^{-1}a=\alpha (b{)}^{-1}b$, then $\alpha (b{a}^{-1})=b{a}^{-1}$, so $a=b$. The displayed map is therefore a bijection. For conjugation by a nonidentity element of a Frobenius complement, it gives the commutator map $y\mapsto h^{-1}{y}^{-1}hy$.

For a Frobenius complement $H$, a class function $\theta$ with $\theta (1)=0$ satisfies $({\theta }^G{)}_H=\theta$. For each nontrivial $\eta \in \mathrm{Irr}(H)$ of degree $d$, the virtual character $(\eta -d{1}_H{)}^G+d{1}_G$ has norm one and positive degree by Frobenius reciprocity. It is therefore irreducible, restricts to $\eta$ and is identically $d$ on the candidate kernel $N$. The intersection of the kernels of these characters is exactly $N$: outside $N$, separation follows from the character of the regular representation of $H$. This proves normality; $N\cap H=1$ and $|N|=[G:H]$ then give the semidirect product.

A Frobenius complement is the nontrivial proper subgroup $H$ in a Frobenius group with the displayed trivial-intersection property. In the action on cosets of $H$, it is a point stabilizer; no nonidentity element fixes two distinct points.

Every finite group whose order is divisible by at most two distinct primes is a solvable group. A nonabelian simple group must therefore have at least three distinct prime divisors.

A nonabelian simple group has no nonidentity conjugacy class of prime-power size. Coprime-degree irreducible characters vanish on such an element: the conjugacy-class sum makes the character-to-degree ratio an algebraic integer, and the Kronecker theorem on algebraic integers in the unit disk makes a nonzero ratio a root of unity, forcing a scalar in a faithful group representation. Column character orthogonality then makes $1/p$ an algebraic integer, impossible. This is the character-theoretic ingredient in Burnside's theorem.

If an abelian subgroup $A$ has index $p^a$, the centralizer of any $1\ne g\in A$ contains $A$, so the class size of $g$ divides $p^a$. The prime-power conjugacy-class obstruction to simplicity rules this out. If $A=1$, the group is a p-group and has nontrivial center of a group, also ruling out nonabelian simplicity.

The order of a finite group is its cardinality, written $|G|$. The identity element is included: $|\mathbb{Z}/6\mathbb{Z}|=6$, and the trivial group has order one, rather than zero.

Keep this count distinct from the order of a group element. In the additive finite cyclic group $\mathbb{Z}/6\mathbb{Z}$, the element $1$ has order six, $2$ has order three, $3$ has order two, and $0$ has order one. Each element's order is the least positive number of repetitions needed to return to the identity element $0$. The order of an element of a finite cyclic group computes these periods directly.

The count also constrains subgroups: Lagrange's theorem gives $|G|=[G:H]|H|$ for a subgroup $H$ of a finite group $G$, where the index of a subgroup counts its cosets. For $H=\{0,2,4\}$ inside $\mathbb{Z}/6\mathbb{Z}$, the two cosets are $\{0,2,4\}$ and $\{1,3,5\}$, so $6=2\cdot 3$. Divisibility is a necessary constraint; it does not imply that every divisor occurs as the order of a group element, as the Klein four-group shows.