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Group axioms and examplesEdexcel A-Level Further Maths: Revision notes

Section 1

Binary operations and closure

A binary operation ∗* on a set SS combines any two elements of SS to give one element. The set SS is closed under ∗* if a∗b∈Sa*b\in S for all a,b∈Sa,b\in S. Example: {1,2,3,4,5}\{1,2,3,4,5\} under multiplication modulo 6 is not closed, because 2×3≡02\times3\equiv0 and 00 is not in the set. Always check closure first: it is the axiom most often failing.

Key termsbinary operationclosure
Common mistake

Assuming closure without testing. Find a single pair whose product leaves the set to show a set is not a group.

Section 2

The four group axioms

A set GG with a binary operation ∗* is a group if:

  1. Closure: a∗b∈Ga*b\in G for all a,b∈Ga,b\in G.
  2. Associativity: (a∗b)∗c=a∗(b∗c)(a*b)*c=a*(b*c) for all a,b,c∈Ga,b,c\in G.
  3. Identity: there is e∈Ge\in G with a∗e=e∗a=aa*e=e*a=a for all a∈Ga\in G.
  4. Inverses: for every a∈Ga\in G there is a−1∈Ga^{-1}\in G with a∗a−1=a−1∗a=ea*a^{-1}=a^{-1}*a=e. The identity is unique, and each element has exactly one inverse. If the operation is also commutative (a∗b=b∗aa*b=b*a for all a,ba,b), the group is called abelian; this is not required by the axioms.
Key termsassociativityidentityinversegroup
Exam tip

Matrix multiplication and composition of functions are already associative, so you can quote that rather than prove it.

Section 3

Examples of groups

  • Integers modulo nn under addition, {0,1,…,n−1}\{0,1,\ldots,n-1\}: identity 00, the inverse of aa is n−an-a (and 00 is self-inverse). This is a group for every nn.
  • Integers modulo nn under multiplication: {1,…,n−1}\{1,\ldots,n-1\} is a group only when nn is prime, e.g. {1,…,6}\{1,\ldots,6\} mod 7; the elements must all be invertible, e.g. {1,3,5,7}\{1,3,5,7\} mod 8 is a group but {1,…,5}\{1,\ldots,5\} mod 6 is not (it is not closed).
  • Non-singular matrices: 2×22\times2 matrices with non-zero determinant under multiplication, with II as identity and the matrix inverse as inverse. Singular matrices have no inverse, so they are excluded.
  • Symmetries of a geometrical figure under composition, e.g. the six symmetries of an equilateral triangle (three rotations including the identity and three reflections).
  • Permutation groups: the permutations of {1,2,3}\{1,2,3\} under composition, a group of order 6. State the order of composition, for example xyxy means do yy first.
Key termsintegers modulo nnon-singular matrixpermutationsymmetry
Common mistake

Including 00 in a multiplicative group modulo nn. 00 has no inverse.

Section 4

Cayley tables

A Cayley table lists all products a∗ba*b for a finite group, with the row element on the left. Use it to check axioms: closure holds if every entry is in the set; the identity is the element whose row and column repeat the headings; and each element appears exactly once in every row and every column (a Latin square), which gives inverses. If the table is symmetrical about the leading diagonal the group is abelian. Example, {1,3,5,7}\{1,3,5,7\} mod 8: the table has 11 down the leading diagonal, since every element is its own inverse.

Key termsCayley table
Exam tip

If an element is repeated in a row or column of a table, the set cannot be a group.

Section 5

Cyclic groups

A group is cyclic if every element is a power (or multiple) of one element gg, called a generator: G={e,g,g2,…}G=\{e,g,g^2,\ldots\}. In additive notation, the elements are the multiples g,2g,3g,…g,2g,3g,\ldots. Examples: {0,1,2,3,4,5}\{0,1,2,3,4,5\} under addition modulo 6 is generated by 11 (and by 55) but not by 22, which produces only 0,2,40,2,4. {1,…,6}\{1,\ldots,6\} under multiplication modulo 7 is generated by 33: 3,2,6,4,5,13,2,6,4,5,1. The group {1,3,5,7}\{1,3,5,7\} mod 8 is not cyclic, because every element squares to the identity.

Key termscyclic groupgenerator
Common mistake

Stopping when you reach the identity early. A generator must produce all the elements before it returns to the identity.

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Exam questions on Group axioms and examples

  1. The set G={1,3,5,7}G=\{1,3,5,7\} under multiplication modulo 8 is a group.
    Determine whether GG is a cyclic group, justifying your answer.2 marks
  2. The set G={1,2,3,4,5,6}G=\{1,2,3,4,5,6\} under multiplication modulo 7 is a group.
    Show that the set {1,2,3,4,5}\{1,2,3,4,5\} under multiplication modulo 6 is not a group.2 marks
  3. Let ee be the identity permutation of {1,2,3}\{1,2,3\}, let rr be the permutation 1→2, 2→3, 3→11\to2,\ 2\to3,\ 3\to1, and let ss be the permutation that swaps 11 and 22 and fixes 33. The product xyxy means apply yy first, then xx. The six permutations of {1,2,3}\{1,2,3\} form a group under this product.
    Find rsrs and srsr, writing each as a permutation, and state what this shows about the group.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).