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Subgroups, Lagrange's theorem and isomorphism questions
A group can contain smaller groups, and their sizes are constrained. Lagrange's theorem states that the order of a subgroup divides the order of the group. Two groups with the same multiplication structure are isomorphic.
7 original questions · 27 marks · the subgroups, lagrange's theorem and isomorphism notes · Further Pure 2
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State Lagrange's theorem and one immediate consequence for the order of an element.
Worked 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.A cyclic group of order 6 is generated by g. List its subgroups and check them against Lagrange's theorem.
Worked 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.A group G has order 8. State the possible orders of its subgroups, and prove that G cannot contain an element of order 3.
Worked 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.Prove that a group of prime order p is cyclic and has no proper subgroups other than the identity.
Worked 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.Decide whether the cyclic group of order 4 and the group {1, 3, 5, 7} under multiplication modulo 8 are isomorphic, with a reason.
Worked 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.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.
Worked 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. All three orders divide 8, as Lagrange requires.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.
Worked 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.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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