Further Mathematics · OCR H245 · Option A · Topic 8.03

Groups

Binary operations, definition of a group, orders of elements and groups, subgroups, cyclic groups, generators, properties of groups and more.

  • 10 sub-topics
  • 2 in Year 12, 8 in Year 13
  • 109 practice questions
  • 10 animated lessons

What the specification says

Every sub-topic OCR 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. 8.03a-bBinary operationsY12
  2. 8.03c-dDefinition of a groupY13
  3. 8.03eOrders of elements and groupsY13
  4. 8.03fSubgroupsY13
  5. 8.03gCyclic groupsY13
  6. 8.03hGeneratorsY13
  7. 8.03i-jProperties of groupsY13
  8. 8.03kLagrange's theoremY13
  9. 8.03lIsomorphismY13
  10. 8.03mAbstract groupsY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 109 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)
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  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.
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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)
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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.
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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)
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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.
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