Maschke's Theorem and Complete Reducibility · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Averaging a representation into pieces

Mathematics · Abstract Algebra · ages 22-23
Name ______________________   Date ____________

  1. Symmetric group on 3 letters. Averaging over the group divides by what number?

    Answer: ______________

  2. Maschke proof averages a projection. What operation makes it respect the group?

    • Average conjugates over the group and divide by the order
    • Delete the projection
    • Multiply by zero
    • Take the determinant

  3. The Maschke proof starts with a projection onto a subrepresentation. What operation makes it respect the group?

    • Delete the projection and start over
    • Average its conjugates over the group and divide by the order
    • Take the determinant of the projection

  4. If the characteristic of the field divides the group order, Maschke's conclusion can fail because one over the order does not exist.

    Circle one:   True   False

  5. You have two irreducible representations and a nonzero linear map between them that respects the group. What must it be?

    • The zero map in disguise
    • A non invertible projection
    • An isomorphism

  6. If the characteristic divides the group order, Maschke can fail because 1 over the order does not exist. Is this correct?

    Circle one:   True   False

  7. Which case shows why division by the group order matters?

    • Cyclic of order 2 over complex numbers
    • Trivial group over any field
    • Cyclic of order 2 in characteristic 2, where 2 equals 0 and division is impossible
    • Group of order 3 over complex numbers

  8. Which case shows why division by the group order matters?

    • The cyclic group of order 2 in characteristic 2, where 2 equals 0
    • The cyclic group of order 2 over the complex numbers
    • The trivial group over any field

  9. A group has order 5 and your field has characteristic 5. Does Maschke's theorem apply?

    • Yes, because 5 is prime
    • No, because 5 equals 0 there and division by the order is impossible
    • Yes, because every field works

  10. Cyclic group of order 3 in characteristic 3. What is 3 modulo 3, which shows division is impossible?

    Answer: ______________

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Answer key

For grown-ups. Fold this page away before handing over the rest.

Averaging a representation into pieces W1-mt_FVNCwUqeTx-s1

  1. 6 · Group has 6 elements, so averaging divides by 6.
  2. Average conjugates over the group and divide by the order · Average conjugates over the group and divide by the order produces an equivariant projection, while the other operations destroy it.
  3. Average its conjugates over the group and divide by the order · Conjugating by each element and averaging spreads the map evenly over the group.
  4. True · The averaging step needs that division, so without it the proof collapses.
  5. An isomorphism · Schur's lemma leaves only two options for such a map, and nonzero rules out zero.
  6. True · Division by zero in the field blocks the averaging step.
  7. Cyclic of order 2 in characteristic 2, where 2 equals 0 and division is impossible · Cyclic of order 2 in characteristic 2, where 2 equals 0 and division is impossible exhibits the failure, while the complex and trivial cases satisfy the hypothesis.
  8. The cyclic group of order 2 in characteristic 2, where 2 equals 0 · In characteristic 2 the divisor 2 equals 0, so normalization is impossible and splitting can fail.
  9. No, because 5 equals 0 there and division by the order is impossible · The characteristic divides the order, so one over the order does not exist.
  10. 0 · 3 leaves remainder 0 modulo 3, so division by 3 is division by 0.
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