The Sylow Theorems and Groups of Small Order · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Counting prime power subgroups

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

  1. A group of order 56 has eight Sylow 7-subgroups. How many elements of order 7 do these subgroups contain altogether?

    Answer: ______________

  2. For a group of order 60, a proposed Sylow 5 count is 11 because 11 is 1 modulo 5. What does the proposal miss?

    • The count must divide 60, and 11 passes that divisibility test.
    • The count must divide 5, and 11 passes that divisibility test.
    • The count must be a multiple of 5.
    • The count must divide 12, and 11 does not.

  3. A group has order 39. Which pair lists all counts of Sylow 3-subgroups permitted by Sylow's divisibility and congruence conditions?

    • 1 or 3
    • 3 or 13
    • 1 or 13

  4. If every Sylow subgroup of a finite group of composite order is normal, then the group is not simple.

    Circle one:   True   False

  5. A group of order 35 has normal Sylow 5-subgroups and normal Sylow 7-subgroups. What follows?

    • It is cyclic, with an element of order 35.
    • It is simple, with an element of order 35.
    • It is cyclic, with an element of order 25.

  6. The cyclic group of order 12 can have four Sylow 3-subgroups because 4 passes the Sylow divisibility and congruence tests.

    Circle one:   True   False

  7. A group of order 12 has four Sylow 3-subgroups, accounting for eight elements of order 3. Why is its Sylow 2-subgroup unique?

    • Every set of four group elements is a subgroup.
    • Every Sylow 2-subgroup equals the four-element complement.
    • The eight order-3 elements form a subgroup, so its complement is a subgroup.
    • Every element outside a Sylow 3-subgroup must have order exactly 2.

  8. For a group of order 50, Sylow counting gives exactly one subgroup of order 25. Why does this rule out simplicity?

    • A subgroup of index 2 makes G abelian.
    • The subgroup of order 25 is normal, nontrivial and proper.
    • A unique subgroup of order 25 makes every subgroup of order 2 normal.

  9. Which pair is the complete list of groups of order 10, up to isomorphism?

    • The cyclic group and the symmetries of a regular 7-gon
    • The cyclic group and the symmetries of a regular 5-gon
    • The cyclic group and the symmetries of a regular 9-gon

  10. A group of order 14 has a of order 7 and b of order 2, with bab^-1=a^-1. How many elements have order 2?

    Answer: ______________

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Answer key

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

Counting prime power subgroups W1-mt_vg2VYkLuz8-s1

  1. 48 · Each subgroup contributes six nonidentity elements. Distinct order-7 subgroups share only the identity, so the total is 8 times 6, or 48.
  2. The count must divide 12, and 11 does not. · Since 60=5 times 12, the count must divide 12 as well as be 1 modulo 5. The proposed count 11 fails divisibility.
  3. 1 or 13 · Since 39=3 times 13, the count divides 13 and is 1 modulo 3. Both divisors, 1 and 13, satisfy the remainder test.
  4. True · With two distinct prime divisors, a normal Sylow subgroup is proper and nontrivial. For composite prime-power order, a central subgroup of order p supplies a proper nontrivial normal subgroup.
  5. It is cyclic, with an element of order 35. · The two prime-order subgroups are cyclic and commute. Generators a and b have a product ab of order 35, which generates the whole group.
  6. False · The arithmetic conditions permit 4 as a candidate, but a cyclic group is abelian. Conjugation fixes its subgroups, so Sylow conjugacy forces its Sylow 3 count to be 1.
  7. Every Sylow 2-subgroup equals the four-element complement. · A subgroup of order 4 contains no element of order 3. Every Sylow 2-subgroup lies in the four-element complement and, having size 4, equals it.
  8. The subgroup of order 25 is normal, nontrivial and proper. · The unique Sylow 5-subgroup is normal. Its order is 25, strictly between 1 and 50, so it is a nontrivial proper normal subgroup.
  9. The cyclic group and the symmetries of a regular 5-gon · For order 2p with p=5, conjugation gives either commuting generators or inversion. These yield the cyclic group and the full pentagon symmetry group.
  10. 7 · The seven distinct elements a^i b square to the identity and are not the identity. The nonidentity elements of <a> instead have order 7.
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