Further Mathematics · AQA 7367 · Option D · Topic DG

Binary operations

Binary operations and groups: Cayley tables, identities, inverses, subgroups, Lagrange's theorem and isomorphisms.

  • 12 sub-topics
  • 6 in Year 12, 6 in Year 13
  • 121 practice questions
  • 12 animated lessons

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. DG1Binary operations, including modular arithmeticY12
  2. DG2CommutativityY12
  3. DG3AssociativityY12
  4. DG4Cayley tablesY12
  5. DG5Identity elementsY12
  6. DG6InversesY12
  7. DG7Language of groups: order, period, subgroupY13
  8. DG8Group axiomsY13
  9. DG9Finite, infinite, cyclic and abelian groupsY13
  10. DG10Lagrange's theoremY13
  11. DG11GeneratorsY13
  12. DG12Isomorphism between finite groupsY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 121 questions in this topic, free, no account needed.

Groups

Not needed for Year 12Easy

Cayley table · Year 13 · 2 marks

  1. Construct the Cayley table for addition modulo 4 on {0, …, 3} and state the identity.
    (2)
Show worked solutionHide worked solution
  1. Each row is a cyclic shift of 0, 1, …
  2. Identity 0.

Total2

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
How did it go?

Report a problem with this question

Groups

Not needed for Year 12Medium

Group axioms · Year 13 · 7 marks

The set GG consists of the integers in {1,2,…,14}\{1, 2, \dots, 14\} that have no factor in common with 15, under multiplication modulo 15.
  1. (a)
    List the elements of GG.
    (1)
  2. (b)
    Show that GG is a group.
    (5)
  3. (c)
    Explain why the set {0,1,2,…,14}\{0, 1, 2, \dots, 14\} is not a group under multiplication modulo 15.
    (1)
Show worked solutionHide worked solution

Part (a)1 mark

  1. {1,2,4,7,8,11,13,14}\{1, 2, 4, 7, 8, 11, 13, 14\}

Part (b)5 marks

  1. Each axiom must be stated and justified.Closure: if aa and bb have no factor in common with 15, neither does ab, and so neither does its remainder.
  2. identity=1\text{identity} = 1
  3. Every element has an inverse in GG.1−11^{-1}=1, 2−1{}= 1, \ 2^{-1}=8, 4−1{}= 8, \ 4^{-1}=4, 7−1{}= 4, \ 7^{-1}=13, 8−1{}= 13, \ 8^{-1}=2, 11−1{}= 2, \ 11^{-1}=11, 13−1{}= 11, \ 13^{-1}=7, 14−1{}= 7, \ 14^{-1}=14{}= 14
  4. Multiplication modulo 15 is associative.

Part (c)1 mark

  1. 0 has no inverse: 0×a≡00 \times a \equiv 0, never 1.

Total7

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
How did it go?

Report a problem with this question

Groups

Not needed for Year 12Hard

Lagrange's theorem · Year 13 · 7 marks

GG is the group {1,2,…,6}\{1, 2, \dots, 6\} under multiplication modulo 7.
  1. (a)
    State the possible orders of the subgroups of GG, giving a reason.
    (2)
  2. (b)
    Show that 3 is a generator of GG.
    (2)
  3. (c)
    Hence find the subgroup of GG of order 3.
    (2)
  4. (d)
    Explain why GG has no subgroup of order 4.
    (1)
Show worked solutionHide worked solution

Part (a)2 marks

  1. 1,2,3,61, 2, 3, 6
  2. By Lagrange's theorem the order of a subgroup divides ∣G∣=6|G| = 6.

Part (b)2 marks

  1. Check the powers, or check gᵈ ≠ 1 for the proper divisors d.3k≢1 for k<6, 36≡13^k \not\equiv 1 \text{ for } k < 6, \ 3^{6} \equiv 1
  2. ord(3)=6\text{ord}(3) = 6

Part (c)2 marks

  1. An element of order 3 is g6/3g^{6/3}.32≡23^{2} \equiv 2
  2. {1,2,4}\{1, 2, 4\}

Part (d)1 mark

  1. 4 does not divide 6.

Total7

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
How did it go?

Report a problem with this question