Linear Representation of a Finite Group

Overview

A representation translates group multiplication into linear algebra. It lets one study symmetry through invariant subspaces, matrices, and traces. The representation need not be faithful, and the field's characteristic determines whether it breaks into simple pieces.

Prerequisites

Know finite groups, vector spaces over a field $F$, and invertible linear maps. For a vector space $V$, the group $\operatorname{GL}(V)$ consists of invertible endomorphisms. A subspace is invariant if every represented group element maps it into itself.

Definition and conventions

A linear representation of a finite group $G$ over $F$ is a homomorphism

Here we focus on finite-dimensional $V$. A nonzero representation is irreducible when its only invariant subspaces are zero and $V$. It is completely reducible if it is a direct sum of irreducible representations.

Two representations are equivalent if an invertible linear map intertwines their actions. The character is $\chi_\rho(g)=\operatorname{tr}(\rho(g))$. The familiar complex character classification requires a suitable setting; it must not be transferred unchanged to arbitrary characteristic.

Intuition

Group elements move vectors while respecting their linear combinations. An invariant subspace is a collection of directions that the action never leaves. Decomposition looks for smaller independent systems of directions whose combined behavior recovers the action.

Examples and non-examples

Let $S_3$ permute the coordinates of $F^3$. The line generated by $(1,1,1)$ is invariant, and so is the plane

If the characteristic is not three, these spaces form a direct sum. Over $\mathbb C$, $W$ is the standard irreducible two-dimensional representation. If the characteristic is three, the displayed line lies in $W$, so that particular decomposition fails.

For a cyclic group of order $p$ over $\mathbb F_p$, assign its generator

Writing $J=I+N$ with $N^2=0$ gives $J^p=I$, so this is a representation. It has an invariant line but no invariant complementary line: the only eigenvectors lie in that line. It is not completely reducible.

Important properties

Maschke's theorem states that if $\operatorname{char}F$ does not divide $|G|$, every finite-dimensional representation is completely reducible. Given an arbitrary projection onto an invariant subspace, average its conjugates by $\rho(g)$ and divide by $|G|$. The resulting projection commutes with the group action, so its kernel is an invariant complement.

This proof explains the characteristic hypothesis directly: the scalar $|G|^{-1}$ must exist. The modular example above shows that the conclusion can fail when it does not.

Connections and further reading

Representations are modules over the group algebra $F[G]$. Over $\mathbb C$, character inner products detect irreducible constituents. See Milne, Chapter 7, Linear representations, Maschke's theorem, and The characters of G.