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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.

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What a learner can do afterwards

  • 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

1 · Read

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.

Try it together

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.

Good to know

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.

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

2 · Watch

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Print a worksheetA4 with an answer key page for grown-ups. No screen, no internet.

Where it sits

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Maschke's Theorem and Complete Reducibility · Mathematics, ages 22 to 23 · LightMySky