Questions on groups and their axioms

Question 1

State the four axioms a set and a binary operation must satisfy to form a group.

Accepted 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.

Question 2

Show that {1, 3, 5, 7} under multiplication modulo 8 is a group, and state the order of each element.

Accepted 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.

Question 3

State what it means for a set to be closed under a binary operation, and give an example of a set and an operation for which closure fails.

Accepted Answer

Closure means that combining any two elements of the set, in either order, gives an element that is itself in the set. B1 for the statement. The odd integers under addition are not closed, since 3 + 5 = 8 and 8 is not odd. B1 for a correct counterexample.

Question 4

The operation * is defined on the integers by a * b = a + b − 3. Find the identity element for *, and find the inverse of 7 under *.

Accepted Answer

The identity e satisfies a * e = a for every a, that is a + e − 3 = a, so e − 3 = 0 and e = 3. Since a * b = b * a here, the same element works on both sides. M1 for setting a * e = a, A1 for e = 3. The inverse x of 7 satisfies 7 * x = 3, that is 7 + x − 3 = 3, so x = −1. B1 for the inverse.

Question 5

Build the Cayley table for {1, 2, 3, 4} under multiplication modulo 5, show the set is a group, and find a generator.

Accepted 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.

Question 6

Explain the difference between the order of a group and the order of an element, using the symmetries of an equilateral triangle.

Accepted 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.

Question 7

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.

Accepted 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.

Question 8

Explain why {0, 1, 2, 3} under multiplication modulo 4 is not a group, and why removing 0 does not repair it.

Accepted 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.

Question 9

The operation * is defined on the real numbers by a * b = a + b − ab. Determine whether * is associative, and determine whether the real numbers form a group under *.

Accepted Answer

For associativity, expand both bracketings. (a * b) * c = (a + b − ab) + c − (a + b − ab)c = a + b + c − ab − ac − bc + abc, while a * (b * c) = a + (b + c − bc) − a(b + c − bc) = a + b + c − bc − ab − ac + abc. The two agree term for term, so * is associative. M1 for expanding one bracketing, A1 for both expansions and the conclusion.
Closure is immediate, since a + b − ab is a real number. For an identity, a * e = a gives a + e − ae = a, that is e(1 − a) = 0 for every a, so e = 0. B1 for the identity.
For an inverse of a, solve a + x − ax = 0, giving x(1 − a) = −a and x = a/(a − 1) provided a ≠ 1. M1 for solving a * x = 0. At a = 1 the operation gives 1 * x = 1 + x − x = 1 whatever x is, so 1 has no inverse and the real numbers do not form a group under *. A1 for the conclusion with 1 named. Removing that single element repairs it: the reals other than 1 do form a group under *.

Question 10

Show that subtraction is not associative on the integers, and state which group axiom this makes the integers under subtraction fail.

Accepted Answer

Test one triple. (1 − 2) − 3 = −1 − 3 = −4, while 1 − (2 − 3) = 1 − (−1) = 2. The two are different, so subtraction is not associative. M1 for testing a specific triple, A1 for the two values. The axiom that fails is associativity, so the integers under subtraction are not a group, even though the set is closed and 0 behaves as an identity on the right. B1 for naming the axiom.

Question 11

Show that the set {1, 2, 4, 5, 7, 8} under multiplication modulo 9 is a cyclic group, naming a generator, and give the order of every element.

Accepted Answer

These are the residues below 9 that share no factor with 9, and multiplying two of them cannot introduce a factor of 3, so the set is closed. Multiplication modulo 9 is associative, 1 is the identity, and the pairings 2 × 5 = 10 ≡ 1, 4 × 7 = 28 ≡ 1 and 8 × 8 = 64 ≡ 1 supply every inverse. B1 for the group check with closure, identity and inverses.
The powers of 2 are 2, 4, 8, 7, 5, 1, which is the whole set, so 2 has order 6 and generates the group, which is therefore cyclic. M1 for the powers of 2, A1 for the generator.
Element orders: 1 has order 1; 2 and 5 have order 6; 4 and 7 have order 3; 8 has order 2. B1 for all six orders. Every one of them divides 6, as Lagrange requires, and the two elements of order 6 are the two generators.

Question 12

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.

Accepted 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. 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.

Question 13

S is the set of 2 × 2 matrices with rows (a, b) and (−b, a), where a and b are real and not both zero. Show that S is closed under matrix multiplication and that every element of S has an inverse in S, and state which results about finite groups have nothing to say about S.

Accepted Answer

Multiplying the matrix with rows (a, b) and (−b, a) by the one with rows (c, d) and (−d, c) gives first row (ac − bd, ad + bc) and second row (−(ad + bc), ac − bd), which is again of the printed shape with a′ = ac − bd and b′ = ad + bc. M1 for the product, A1 for it having the same shape. It is a member of S provided a′ and b′ are not both zero, and a′² + b′² = (a² + b²)(c² + d²), a product of two positive numbers. B1 for that.
The determinant is a² + b², which is positive, so no element of S is singular, and the inverse is (1/(a² + b²)) times the matrix with rows (a, −b) and (b, a), which is the printed shape with a/(a² + b²) and −b/(a² + b²) in place of a and b. M1 for the determinant, A1 for the inverse lying in S.
S is infinite, so it has no order as a number, and Lagrange's theorem, the statement that every element order divides the order of the group, and the counting of groups of a given order all say nothing here. Elements need not have finite order either: the powers of 2I are 2ⁿI, never the identity. B1 for naming what stops applying, with a reason. The map sending a + ib to this matrix turns S into the non-zero complex numbers under multiplication, which is where the product rule above comes from.

Question 14

G is a group. Prove from the axioms that G has exactly one identity element, and that each element of G has exactly one inverse.

Accepted Answer

Suppose e and f are both identities of G. Since e is an identity, ef = f; since f is an identity, ef = e. Hence e = f, so there is only one. M1 for forming the product ef and reading it both ways, A1 for the conclusion.
Now let a be an element of G and suppose b and c are both inverses of a, so ab = ba = e and ac = ca = e. Then b = be = b(ac) = (ba)c = ec = c. M1 for inserting the identity as ac, M1 for using associativity to regroup, A1 for b = c.

Question 15

The operation * is defined on the open interval −1 < x < 1 by a * b = (a + b)/(1 + ab), and may be assumed associative. Show that the interval is closed under *, that * has an identity, and that every element has an inverse, and write down the inverse of ½.

Accepted Answer

For a and b in the interval, ab > −1, so 1 + ab > 0 and the quotient is defined. M1 for the denominator never vanishing.
For closure, compare the squares: (1 + ab)² − (a + b)² = 1 + 2ab + a²b² − a² − 2ab − b² = (1 − a²)(1 − b²), which is positive because both brackets are. M1 for the identity, A1 for concluding |a * b| < 1, so a * b lies in the interval.
For the identity element, (a + e)/(1 + ae) = a gives a + e = a + a²e, so e(1 − a²) = 0 for every a in the interval and e = 0. B1. For an inverse, (a + x)/(1 + ax) = 0 needs a + x = 0, so x = −a, which lies in the interval whenever a does. B1. Hence the inverse of ½ is −½. B1. The interval is a group under * although it has no largest and no smallest element, and although the operation is not addition of any kind: this is how velocities combine in special relativity, with 1 standing for the speed of light.