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Groups and their axioms questions
Four axioms about a set and a binary operation, closure, associativity, an identity and inverses, define a group. Symmetries of a shape, arithmetic modulo n and sets of matrices all satisfy them, which is why one set of results applies to all of them.
7 original questions · 28 marks · the groups and their axioms notes · Further Pure 2
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State the four axioms a set and a binary operation must satisfy to form a group.
Worked answer
Closure: the operation on two elements of the set gives an element of the set. Associativity: (ab)c = a(bc) for all elements. Identity: there is an element e with ea = ae = a for every a. Inverses: every a has an element a⁻¹ in the set with aa⁻¹ = a⁻¹a = e. B1 for closure and associativity, B1 for the identity, B1 for inverses. Commutativity is not required.Show that {1, 3, 5, 7} under multiplication modulo 8 is a group, and state the order of each element.
Worked answer
The products are 3 × 5 = 15 ≡ 7, 3 × 7 = 21 ≡ 5 and 5 × 7 = 35 ≡ 3, and each element squares to 1, so the set is closed. Multiplication modulo 8 is associative, 1 is the identity, and every element is its own inverse. The identity has order 1; 3, 5 and 7 each have order 2. M1 for the closure check, A1 for the remaining axioms, B1 B1 for the orders.Build the Cayley table for {1, 2, 3, 4} under multiplication modulo 5, show the set is a group, and find a generator.
Worked answer
Rows for 1, 2, 3, 4 read (1, 2, 3, 4), (2, 4, 1, 3), (3, 1, 4, 2) and (4, 3, 2, 1). Every entry lies in the set, the identity 1 appears exactly once in each row and column so inverses exist, and modular multiplication is associative. Powers of 2 give 2, 4, 3, 1, so 2 has order 4 and generates the whole group, so it is cyclic. (3 works as well; 4 has order 2 and does not.) M1 A1 for the table, M1 for the group check, A1 for the conclusion, B1 for the generator.Explain the difference between the order of a group and the order of an element, using the symmetries of an equilateral triangle.
Worked answer
The order of a group is the number of elements it contains; the order of an element is the least positive power of it that gives the identity. The symmetry group of an equilateral triangle has order 6, three rotations and three reflections. The rotation through 120° has order 3, since three of them return the triangle to its starting position, while each reflection has order 2. B1 for the order of a group, B1 for the order of an element, B1 for the triangle example.Show that the set of matrices with rows (1, n) and (0, 1), for integer n, forms a group under matrix multiplication, and say which familiar group it matches.
Worked answer
Multiplying the matrix for m by the matrix for n gives rows (1, m + n) and (0, 1), so the set is closed and the operation behaves like adding the integers. Matrix multiplication is associative, n = 0 gives the identity matrix, and the matrix for −n inverts the matrix for n. Since the operation is addition of the n values, the group is isomorphic to the integers under addition. M1 A1 for the product and closure, B1 for the identity and inverses, B1 for the isomorphism.Explain why {0, 1, 2, 3} under multiplication modulo 4 is not a group, and why removing 0 does not repair it.
Worked answer
With 0 included there is no inverse for 0, since no residue multiplied by 0 gives the identity 1. Removing it leaves {1, 2, 3}, but 2 × 2 = 4 ≡ 0, which is outside the set, so closure fails and 2 has no inverse either. The multiplicative residues modulo n form a group only when the residues kept are those coprime to n, which for 4 means {1, 3}. B1 for 0 having no inverse, M1 for testing closure on the remaining set, A1 for 2 × 2 ≡ 0.G is a group in which every element x satisfies x² = e. Prove that G is abelian, and state a group of order 4 with this property.
Worked answer
Every element is its own inverse, since x² = e gives x⁻¹ = x. Take any a and b in G. Then ab lies in G, so (ab)² = e and ab is its own inverse, that is ab = (ab)⁻¹. But (ab)⁻¹ = b⁻¹a⁻¹ = ba, using the order reversal for the inverse of a product and then the fact that each element is self-inverse. Hence ab = ba for all a and b, so G is abelian. {1, 3, 5, 7} under multiplication modulo 8 has order 4 and every element squares to 1, so it is an example. The whole proof turns on applying the hypothesis to the product ab rather than to a and b separately. B1 for every element being self-inverse, M1 for applying the hypothesis to ab, A1 for the reversal of the inverse, A1 for ab = ba, A1 for the conclusion, B1 for the example.
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