---
title: "Maschke's Theorem and Complete Reducibility"
description: "Prove that over a field of the right characteristic every representation splits into irreducible pieces, and see exactly where the proof needs that hypothesis."
canonical: https://lightmysky.com/learn/mathematics/maschkes-theorem-and-complete-reducibility-mt_FVNCwUqeTx
source: https://lightmysky.com/learn/mathematics/maschkes-theorem-and-complete-reducibility-mt_FVNCwUqeTx.md
retrieved: 2026-09-12
---

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# Maschke's Theorem and Complete Reducibility

Prove that over a field of the right characteristic every representation splits into irreducible pieces, and see exactly where the proof needs that hypothesis.

Subject: Mathematics · Area: Abstract Algebra · Ages 22 to 23
Page: https://lightmysky.com/learn/mathematics/maschkes-theorem-and-complete-reducibility-mt_FVNCwUqeTx

## Ready when they can

- Prove Maschke's theorem by averaging a projection over the group
- Point at the division by the group order and give a case where the theorem fails
- Apply Schur's lemma to constrain maps between irreducible representations

## Lesson: Averaging a representation into pieces

Maschke's theorem says that over a field of the right characteristic, every representation of a finite group splits into irreducible pieces. The proof starts with any projection onto a subrepresentation and forces it to respect the group. You conjugate the projection by each group element and average the results: the averaged map is equivariant, and its kernel and image split the space into stable complements.

**Example.** For the symmetric group on 3 letters, averaging runs over 6 elements, so the sum is divided by 6. A smaller cousin shows the arithmetic: with a group of 3 elements, a sum of conjugated projections with diagonal entries 6 and 6 becomes 2 and 2 after dividing each entry by 3. Division by the group order is what normalizes the sum into a genuine projection.

That division is exactly where the hypothesis lives: one over the group order must exist in the field. When the characteristic divides the group order, the step collapses. The classic failure is the cyclic group of order 2 in characteristic 2, where 2 equals 0 and division is impossible. Over the complex numbers the order is never zero, so every representation splits as a direct sum of irreducibles.

**Tip.** Once a representation splits, Schur's lemma constrains the maps between the pieces: any nonzero map between irreducible representations is an isomorphism. Use it to read off which pieces repeat and how they pair up. If a map between irreducibles is neither zero nor invertible, recheck your irreducibility claim first.

**Recap.** Average conjugates, divide by the order where it exists, and split every representation into irreducibles.

## Practice

17 questions on this page, each with its working shown.

## Needs first

- [Rings, Fields and Their First Properties](https://lightmysky.com/learn/mathematics/rings-fields-and-their-first-properties-mt_BmhIook13o)
- [Group Representations and the Group Algebra](https://lightmysky.com/learn/mathematics/group-representations-and-the-group-algebra-mt_y5JlvQGKYW)

## Opens up

- [Characters and the Orthogonality Relations](https://lightmysky.com/learn/mathematics/characters-and-the-orthogonality-relations-mt_RleR7YGXjX)