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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.

Suppose every nonidentity element of a finite group $K$ is either a pi-element or an element for complementary primes. An elementary subgroup cannot have nontrivial factors of both kinds, since the product of nonidentity commuting coprime-order elements would have mixed prime order. Choose an integer $d$ congruent to $1$ modulo $|K{|}_{\pi }$ and to $0$ modulo $|K{|}_{{\pi }^{\prime }}$. Set $f(1)=d$, $f=1$ on nonidentity pi-elements and $f=0$ on the other nonidentity elements. On each elementary subgroup this is an integer combination of the trivial and regular characters. Brauer's characterization of characters supplies the asserted generalized character.

Under the invariant, coprime-degree and determinant hypotheses of the coprime-degree determinant extension theorem, without assuming $G/N$ solvable, each cyclic-quotient subgroup $N⟨g⟩$ has a unique determinant-normalized extension. Use its value at $g$ to define $f$. Uniqueness makes $f$ conjugacy-invariant. On an elementary subgroup $E$, it agrees with the restriction of the extension to $NE$, since $NE/N$ is a solvable quotient of $E$. Brauer's characterization of characters therefore makes $f$ a generalized character, with $f_N=\theta$.

Let $\mathrm{Irr}(G)$ denote the irreducible characters of finite-dimensional complex group representations of $G$. A generalized character, or virtual character, is an integral linear combination of these irreducible characters. A class function is a function constant on conjugacy classes. For class functions use the character inner productA p-elementary group is a direct product of groups $P\times C$, where $P$ is a p-group and $C$ is a cyclic group of order coprime to $p$. An elementary subgroup is a subgroup of this form for some prime $p$. Either factor may be trivial. This notion is different from an elementary abelian group.

Brauer's characterization of characters says that a complex class function $f$ on $G$ is a generalized character exactly when its restriction to every elementary subgroup is a generalized character there. Equivalently,The character orthogonality relations make these inner products the coefficients of the restrictions in their irreducible bases. The theorem detects the integral character lattice. An ordinary character of a representation additionally requires every global irreducible coefficient $⟨f,\chi {⟩}_G$ to be nonnegative; integrality alone characterizes virtual characters, not necessarily actual representations.

If $A$ is empty, take the zero generalized character; if $B$ is empty, take the trivial character. Otherwise let $\pi$ be the set of primes dividing the orders of elements of $A$. The coprimality hypothesis makes every element of $B$ a ${\pi }^{\prime }$-element. Thus every nonidentity element of $G$ is either a pi-element in $A$ or a complementary-prime element in $B$. In particular these sets are unions of conjugacy classes, since conjugation preserves element order.

Put $m=|G{|}_{\pi }$ and $n=|G{|}_{{\pi }^{\prime }}$. Their greatest common divisor is one, so the Chinese remainder theorem supplies an integer $d$ such thatDefine the class function $f$ by $f(1)=d$, $f=1$ on $A$ and $f=0$ on $B$. We verify Brauer's characterization of characters directly.

For an elementary subgroup $E=P\times C$, the cyclic factor can be split into its $\pi$-part and ${\pi }^{\prime }$-part. Attach $P$ to the part containing its prime. This expresses $E=E_{\pi }\times E_{{\pi }^{\prime }}$. Both factors cannot be nontrivial: choosing a nonidentity element in each would give a commuting product of mixed prime order, whereas every element of $G$ has order involving only one of the two prime sets. Hence $E$ is wholly a $\pi$-group or wholly a ${\pi }^{\prime }$-group.

Let $r_E$ be the character of the regular representation of $E$, with value $|E|$ at the identity and zero elsewhere. In the two cases respectively,The coefficients are integers because $|E|$ divides $m$ or $n$. Thus both are generalized characters. The trivial subgroup causes no difficulty since $d$ is an integer. By Brauer's characterization of characters, $f$ is a generalized character of $G$.

This is a generalized character separating complementary prime elements; the freely chosen value at the identity is what makes all elementary restrictions integral.