Questions on subgroups, Lagrange's theorem and isomorphism

Question 1

State Lagrange's theorem and one immediate consequence for the order of an element.

Accepted Answer

The order of any subgroup of a finite group divides the order of the group. Since the powers of a single element form a subgroup, the order of every element divides the order of the group as well. B1 for the theorem, B1 for the order of an element dividing the order of the group.

Question 2

A cyclic group of order 6 is generated by g. List its subgroups and check them against Lagrange's theorem.

Accepted Answer

The subgroups are {e}, {e, g³}, {e, g², g⁴} and the whole group, of orders 1, 2, 3 and 6. Each of those divides 6, as Lagrange requires, and every divisor of 6 is achieved: that always happens in a cyclic group, though not in general. B1 for {e, g³}, B1 for {e, g², g⁴} with the two trivial subgroups, B1 for the orders dividing 6.

Question 3

State what has to be checked to show that a non-empty subset H of a group G is a subgroup of G.

Accepted Answer

H must be closed under the operation of G, so that xy lies in H whenever x and y do, and closed under inverses, so that x⁻¹ lies in H whenever x does. B1 for closure under the operation, B1 for closure under inverses. Associativity is inherited from G and needs no check, and the identity comes free: H is non-empty, so it holds some x, and then x⁻¹ and hence xx⁻¹ = e. For a finite H closure alone is enough, since the powers of any element eventually return to the identity.

Question 4

The integers modulo 12 form a cyclic group of order 12 under addition. List all its subgroups, giving the order of each, and check them against Lagrange's theorem.

Accepted Answer

A cyclic group of order 12 has exactly one subgroup for each divisor of 12, generated by that many steps round the cycle. The divisors are 1, 2, 3, 4, 6 and 12, so the subgroups are {0}; {0, 6}; {0, 4, 8}; {0, 3, 6, 9}; {0, 2, 4, 6, 8, 10}; and the whole group. B1 for the divisors of 12, B1 B1 for the six subgroups listed with their orders 1, 2, 3, 4, 6 and 12. Every one of those orders divides 12, as Lagrange requires. A cyclic group reaches every divisor, which is not true of groups in general: Lagrange says which orders are possible, never that each possible order occurs.

Question 5

A group G has order 8. State the possible orders of its subgroups, and prove that G cannot contain an element of order 3.

Accepted Answer

Subgroup orders must divide 8, so they are 1, 2, 4 or 8. If some element a had order 3 then {e, a, a²} would be a subgroup of order 3, and 3 does not divide 8. That contradicts Lagrange, so no such element exists. B1 for the possible orders 1, 2, 4 and 8, M1 for constructing {e, a, a²}, A1 for its order being 3, A1 for the contradiction with Lagrange.

Question 6

Prove that a group of prime order p is cyclic and has no proper subgroups other than the identity.

Accepted Answer

By Lagrange any subgroup has order dividing p, so its order is 1 or p: the only subgroups are the identity and the whole group. Take any element a other than the identity. The subgroup generated by a is not {e}, so it must be the whole group, and therefore a generates G. Hence G is cyclic, and every non-identity element is a generator. M1 for applying Lagrange, A1 for the only subgroup orders being 1 and p, M1 for taking a non-identity element, A1 for concluding that G is cyclic.

Question 7

Decide whether the cyclic group of order 4 and the group {1, 3, 5, 7} under multiplication modulo 8 are isomorphic, with a reason.

Accepted Answer

They are not. The cyclic group of order 4 contains an element of order 4, its generator. In {1, 3, 5, 7} every element squares to 1, so no element has order more than 2. An isomorphism preserves the order of every element, so no bijection between them can be one. B1 for an element of order 4 in the cyclic group, B1 for every element of {1, 3, 5, 7} squaring to 1, B1 for isomorphism preserving orders.

Question 8

The symmetries of a square form a group of order 8. Find a subgroup of order 4 in two genuinely different ways, and one of order 2.

Accepted Answer

The four rotations, through 0°, 90°, 180° and 270°, form a cyclic subgroup of order 4, generated by the quarter turn.
The identity, the rotation through 180° and the two reflections in the lines through opposite edge midpoints form a different subgroup of order 4. It is closed, since the product of those two reflections is the half turn, and every element squares to the identity, so it has no element of order 4 and is not isomorphic to the first.
The identity with the rotation through 180° gives a subgroup of order 2. B1 for the rotation subgroup, M1 A1 for a second subgroup of order 4 with its closure, B1 for the subgroup of order 2.

Question 9

The non-zero rational numbers form a group G under multiplication. Decide, with reasons, whether each of these subsets is a subgroup of G: (i) the positive rationals; (ii) the numbers 2ⁿ for integer n; (iii) the rationals greater than 1.

Accepted Answer

Test each for closure under multiplication and under inverses. B1 for using that test.
(i) A product of two positive rationals is a positive rational, and the reciprocal of a positive rational is positive, so this is a subgroup. B1.
(ii) 2^m × 2ⁿ = 2^m+n and the inverse of 2ⁿ is 2⁻ⁿ, both of the same form, so this is a subgroup, and it is infinite and cyclic with generator 2. B1.
(iii) This is not a subgroup. The identity 1 is not greater than 1, so the subset does not contain it, and the inverse of 3 is 1/3, which is not greater than 1 either. B1.
The subset in (iii) is closed under multiplication, which is why closure on its own settles nothing for an infinite subset: the inverse test is what separates it from the other two.

Question 10

The symmetries of an equilateral triangle form a group of order 6, generated by a rotation r through 120° and a reflection m. Show that the three rotations form a subgroup, decide whether {e, r, m} is a subgroup, and explain why the group has no subgroup of order 4.

Accepted Answer

The rotations are e, r and r². Composing any two of them turns the triangle through a multiple of 120°, which is again one of the three, so the set is closed; each has an inverse among them, since r and r² undo each other. That makes a subgroup of order 3, and 3 divides 6. M1 for the closure check, A1 for the subgroup of order 3.
{e, r, m} is not a subgroup: r followed by m is a reflection in a different line, which is not one of the three elements listed, so closure fails. B1.
A subgroup of order 4 would need 4 to divide 6, and it does not, so by Lagrange's theorem no such subgroup exists. B1. Notice that {e, r, m} has order 3, a divisor of 6, and is still not a subgroup: Lagrange rules orders out, and never rules a subset in.

Question 11

H and K are subgroups of a group G. Prove that the set of elements lying in both H and K is itself a subgroup of G.

Accepted Answer

Write J for the set of elements belonging to both. J is non-empty, because the identity e lies in H and in K, both being subgroups, so e lies in J. M1 for the identity.
Closure: take x and y in J. Then x and y both lie in H, so xy lies in H; and they both lie in K, so xy lies in K. Hence xy lies in J. M1 for applying closure in each subgroup separately.
Inverses: take x in J. Then x⁻¹ lies in H and x⁻¹ lies in K, so x⁻¹ lies in J. A1 for inverses. Associativity is inherited from G, so J satisfies every requirement and is a subgroup. A1 for the conclusion.
The same argument fails for the set of elements lying in H or in K: in the symmetries of a triangle, two different subgroups of order 2 have a union of three elements, which is not closed.

Question 12

The group D of the symmetries of a square has order 8. The set U = {1, 3, 5, 7, 9, 11, 13, 15} forms a group of order 8 under multiplication modulo 16. Determine, with full justification, whether D and U are isomorphic.

Accepted Answer

Matching orders is not enough. An isomorphism preserves the order of every element, so compare how many elements of each order the two groups hold.
In U. 7² = 49 ≡ 1, 9² = 81 ≡ 1 and 15² = 225 ≡ 1, so 7, 9 and 15 have order 2. For 3: 3² = 9, 3³ = 27 ≡ 11, 3⁴ = 81 ≡ 1, so 3 has order 4, and the same for 5, 11 and 13. The tally is one of order 1, three of order 2, four of order 4.
In D. The quarter turns, through 90° and 270°, have order 4. The half turn has order 2, and so does each of the four reflections, since reflecting twice in the same line does nothing. The tally is one of order 1, five of order 2, two of order 4.
The two lists differ, so D and U are not isomorphic.
A second argument settles it alone. U is a group of residues under multiplication, so it is abelian. D is not: rotating a square through 90° and then reflecting gives a different symmetry from reflecting first. An isomorphism carries a commuting pair to a commuting pair, so an abelian group cannot be isomorphic to a non-abelian one.
M1 for finding element orders in U, A1 for the tally in U, M1 for the element orders in D, A1 for the tally in D, A1 for the conclusion, B1 for U being abelian, B1 for D not being abelian. Lagrange rules nothing out here. Both groups have all their element orders in {1, 2, 4}, every one of which divides 8. Lagrange says which orders are possible, never how many elements take each.

Question 13

G is a group. Prove the cancellation law, that ax = ay implies x = y for all elements a, x and y of G, and deduce that every row of the Cayley table of G contains each element of G exactly once.

Accepted Answer

a has an inverse a⁻¹ in G, so multiply both sides on the left by it: a⁻¹(ax) = a⁻¹(ay). M1 for multiplying by the inverse on the correct side. By associativity that is (a⁻¹a)x = (a⁻¹a)y, that is ex = ey, so x = y. M1 for using associativity to regroup, A1 for the conclusion.
The row of the table labelled a lists the products ax as x runs through G. Every entry lies in G by closure. If two entries in that row were equal, say ax = ay with x and y different, cancellation would force x = y, a contradiction, so the entries are all different. M1 for applying cancellation to two entries of one row. There are as many entries as elements of G and no two agree, so each element appears exactly once. A1.
The same argument on the right gives the columns, so the table is a Latin square. Multiplying on the left is required in the first step: without commutativity, multiplying ax = ay on the right by a⁻¹ gives axa⁻¹ = aya⁻¹ and settles nothing.

Question 14

State how many groups there are, up to isomorphism, of each order from 1 to 7. A group G has order 6 and is not abelian; identify G, giving a reason.

Accepted Answer

The counts are 1, 1, 1, 2, 1, 2, 1 for the orders 1 to 7 in turn. B1 for the seven numbers. Orders 2, 3, 5 and 7 are prime, and a group of prime order is cyclic, so there is exactly one of each; order 4 and order 6 each carry two. B1 for prime order forcing a single cyclic group. The two groups of order 6 are the cyclic group and the group of symmetries of an equilateral triangle. The cyclic group is abelian, since its elements are powers of one generator and powers of a single element commute. M1 for ruling the cyclic group out on those grounds. So G must be the symmetry group of the equilateral triangle, three rotations and three reflections. A1. It really is non-abelian: rotating through 120° and then reflecting is a different symmetry from reflecting and then rotating. B1 for confirming that G exists and is non-abelian, rather than leaving the identification unchecked.

Question 15

G is a group of order 4. Prove that either G is cyclic or every element of G other than the identity has order 2, and deduce that there are exactly two groups of order 4 up to isomorphism.

Accepted Answer

By Lagrange's theorem the order of any element divides 4, so it is 1, 2 or 4, and only the identity has order 1. M1 for Lagrange, A1 for the possible element orders.
If some element a has order 4 then e, a, a², a³ are four distinct elements, which is all of G, so G is cyclic and generated by a. M1 for that case.
Otherwise every non-identity element has order 2, so G = {e, a, b, c} with a² = b² = c² = e, each element being its own inverse. Consider ab. It cannot be e, since that would make b = a⁻¹ = a; it cannot be a, since that would make b = e; and it cannot be b, since that would make a = e. So ab = c, and the same argument fixes every other product: the table is forced, and the group is determined. M1 for eliminating the three impossible values of ab, A1 for ab = c and the table being forced.
So there are at most two groups of order 4, and there are at least two, since the cyclic one has an element of order 4 and the other does not, and an isomorphism preserves the order of every element. Hence exactly two. B1 for the deduction with that reason. The second group is the one the residues {1, 3, 5, 7} under multiplication modulo 8 belong to.