MATH 402A - Solutions for the suggested problems. A. (Groups of order 8. (a Which of the five groups G (as specified in the question have the following property: G has a normal subgroup N such that N = Z/2Z and G/N = Z/2Z Z/2Z? Answer: Among the abelian groups, if G = Z/2Z Z/2Z Z/2Z or if G = Z/4Z Z/2Z, then G has the stated property, as is easily verified. However, if G is cyclic, then one of the homework problems shows that every quotient group of G is also cyclic. Therefore G does not have the stated property. If G is nonabelian and has order 8, then G has the stated property. Thus, both Q 8 and D 8 have the stated property. We will justify this statement. One helpful observation is that if G is nonabelian, then G/Z(G cannot be cyclic. This is because of the following result which was proved in class. Lemma. If G is any group and G/Z(G is cyclic, then G is abelian. It follows that the order of G/Z(G must be at least 4 if G is nonabelian. Now if G = Q 8 or G = D 8, one can verify that Z(G = 2. Hence Z(G = Z/2Z. Also, G/Z(G has order 4 and is not cyclic. Hence G/Z(G = Z/2Z Z/2Z. Combining these observations, if we let N = Z(G, then the above stated property holds. (b Which of the five specified groups G have the following property: G has a normal subgroup N such that N = Z/2Z and G/N = Z/4Z? Answer: Only two of the five groups have this property. We must have N Z(G. (See question E from problem set 4. If N = Z(G, then G/Z(G will be cyclic. Hence, the lemma stated above shows that G must be abelian (in which case we actually have Z(G = G. Thus, G must be isomorphic to one of the following two groups (up to isomorphism Z/8Z or Z/4Z Z/2Z. since Z/2Z Z/2Z Z/2Z clearly doesn t have the stated property. The above two groups do have the stated property, as is easily verified. (c Which of the five groups G have the following property: G has a normal subgroup N such that N = Z/2Z Z/2Z and G/N = Z/2Z?
Answer: It suffices to have a subgroup N of G isomorphic to Z/2Z Z/2Z. Such a subgroup would have index 2 and hence would automatically be a normal subgroup of G. The corresponding quotient group would have order 2 and would then be isomorphic to Z/2Z. Groups of order 8 having such a subgroup are isomorphic to Z/2Z Z/2Z Z/2Z, Z/4Z Z/2Z, or D 8. Now Q 8 has only one element of order 2 and therefore cannot contain any subgroups isomorphic to Z/2Z Z/2Z. Also, Z/8Z is cyclic and hence does not contain a subgroup isomorphic to the group Z/2Z Z/2Z because that group is not cyclic. (d Determine all the conjugacy classes in each of the five groups of order 8. Answer: For the three abelian groups, the conjugacy classes are just the singletons consisting of the individual elements. For the quaternionic group Q 8, the conjugacy classes are: {1}, { 1}, {i, i}, {j, j}, {k, k}. For D 8, the conjugacy classes are: {i}, {r 2 }, {r, r 3 }, {s, sr 2 }, {sr, sr 3 } where r = (1 2 3 4 (which corresponds to a 90 o rotation of a square with vertices labeled 1, 2, 3, 4 and s = (12(34 (which corresponds to one of the reflections of that square. B. Suppose that G is a finite group and that G contains an abelian subgroup A such that [G : A] = 2. (a Suppose that a A. Prove that the conjugacy class of a in G has cardinality equal to 1 or 2. Proof. The cardinality of the conjugacy class for a is equal to the index [G : H], where H = C G (a, the centralizer of a in G. Obviously, since A is abelian and a A, we have A H G. Thus, by Lagrange s theorem, A divides H and so H = t A for some positive integer t. But, H divides G = 2 A and so we have t = 1 or 2. Therefore, the cardinality of the conjugacy
class of a in G is G / H = 2 A /t A = 2/t, which is either 1 or 2. Notice that we have H = A or H = G. In the first case, the conjugacy class of a has cardinality 2. In the second case, that cardinality is 1. (b Suppose that b G, but b A. Show that the cardinality of the conjugacy class of b in G divides A. Give an example of a group G and a subgroup A where the conjugacy class of any such b has cardinality equal to A. Solution: Note that A is a normal subgroup of G since [G : A] = 2, G/A has order 2, and ba is a nontrivial element of the group G/A. Hence ba has order 2 in G/A. Suppose that m is the order of b. Then b m = e, the identity element of G. Hence (ba m = b m A = A and so the order of ba in the quotient group G/A divides m. That is, m is divisible by 2. Now, let H = C G (b, the centralizer of b in G. Note that H contains the cyclic subgroup b, which has order m, and hence m divides H. Thus, H is even. The cardinality of the conjugacy class of b is equal to G / H. Since H is even, G / H divides 1 2 G. However A has index 2 in G and hence A = 1 2 G. Therefore, the cardinality of the conjugacy class of b in G divides A, as stated. As an example, let G = D 10. Then G = 10 and G is nonabelian. Let A be the subgroup of G consisting of rotations, which is a cyclic subgroup of order 5. If b G, but b A, then b is a reflection and has order 2. The number of conjugates of b in G divides G / b = 5. However, since G is a product of two primes and G is nonabelian, we know that Z(G is trivial. (This uses homework problem 4 on page 95. Hence b Z(G. Hence the cardinality of the conjugacy class of G is 5, which is indeed equal to A. C. Prove that if G is a group of order 35, then G = Z/35Z. Proof. First of all, suppose that G is a group and G = 35. We will prove that G has at least one element of order 5 and at least one element of order 7.. If G is abelian, then we have already proved this. (It is the special case of Cauchy s theorem for finite abelian groups, which we proved one day in class. So now assume that G is a nonabelian group of order 35. According to one of the homework problems (page 95, problem 4, we have Z(G = {e}, where e is the identity element in G. Thus, if a G and a e, then a Z(G. Hence the conjugacy class of a in G has cardinality > 1. That cardinality must divide G / a = 35/ a. Therefore, if a = 5, then the conjugacy class of a has cardinality 7. If a = 7, then the conjugacy class of a has cardinality 5. Let x denote the number of distinct conjugacy classes in G for elements of order 5 and let y denote the number of distinct conjugacy classes in G for elements of order 7. Then we have 7x + 5y = G 1 = 34. This equation implies that the nonnegative integers x and y must actually be positive. Therefore, G indeed must contain elements of order 5 and elements of order 7. Thus, we have proved that if G
is a group (abelian or nonabelian and G = 35, then G contains elements of order 5 and of order 7. Let b be an element of G of order 7 and let N = b. Thus, N is a subgroup of G of order 7. According to problem C in problem set 5, it follows that N must be a normal subgroup of G. We will prove that N Z(G. To prove this, suppose that a N, a e. Then a = 7. Then the the conjugacy class of a in G must be contained in N. As explained in class, the cardinality of that conjugacy class divides G / a = 5. Hence the cardinality of that conjugacy class is 1 or 5. Since N has six elements, apart from e, it is clear that at least one of those elements a has just one conjugate. This means that N contains an element a of order 7 such that a Z(G. Hence Z(G {e}. Using problem4 on page 95 again, it follows that G is abelian. We have proved that if G is a group of order 35, then G is abelian. To complete the proof, we now know that G is abelian and that G contains an element b of order 7 and an element c of order 5. (We are using the special case of Cauchy s theorem again. We have bc = cb. Note that (bc 5 = b 5 c 5 = b 5 e, (bc 7 = b 7 c 7 = c 7 = c 2 e and so the order of bc cannot divide 5 or 7. It must divide 35. Hence bc has order 35. It follows that bc is a subgroup of G and has order 35, as does G itself. Therefore, G bc, and so G is indeed a cyclic group. D. Let A = Z/2Z Z/2Z. For each of the following groups G, determine if G has a subgroup isomorphic to A. Justify your answers fully. The group G = S 3 has no subgroup isomorphic to A. Justification: Since G = 6, any subgroup of G has order dividing 6. Hence G cannot have a subgroup of order 4. But A = 4 and any group isomorphic to A must have order 4. The group G = S 4 has a subgroup isomorphic to A, namely the Klein 4-group K = { i, (12(34, (13(24, (14(23 } We know that K is a subgroup of S 4, that K = 4, and that none of the elements of K has order 4. Hence K is not cyclic. Hence K is isomorphic to A, using a theorem proved in class about groups of order 4. (We actually proved a theorem describing the isomorphism classes of groups of order p 2, where p is any prime. The quaternionic group G = Q 8 has no subgroup isomorphic to A. In fact, G has only one element of order 2, namely 1. Therefore, if H is a subgroup of G of order 4, then H must contain an element of order 4. Thus, H must be a cyclic group of order 4 and cannot be isomorphic to A.
The group G = D 8 has a subgroup isomorphic to A. Using the definition given in class, we can regard D 8 as a subgroup of S 4. That subgroup contains K. As pointed out above, K is isomorphic to A. Hence D 8 contains a subgroup isomorphic to A. The group G = Z/4Z Z/2Z contains a subgroup isomorphic to A. We will use additive notation for G. Since G is abelian, the map φ : G G defined by φ(g = 2g for all g G is a homomorphism. Let H = Ker(φ = { g G 2g = e } where e is the identity element of G. Of course, K is a subgroup of G. We will use the following notation. If a Z, we let [a] 2 = a+2z, which is an element of Z/2Z. We let [a] 4 = a+4z, which is an element of Z/4Z. Using this notation for congruence classes, this subgroup H can be described as follows: H = { ( [0] 4, [0] 2, ( [0]4, [1] 2, ( [2]4, [0] 2, ( [2]4, [1] 2 } Since H has order 4 and every element of H has order 1 or 2, we know that H is isomorphic to A. The group G = Z/48Z has no subgroup isomorphic to A. Justification: Since G is cyclic, every subgroup of G must also be cyclic. But A is not cyclic. Any group isomorphic to A will also fail to be cyclic. Hence no subgroup of G can be isomorphic to A. E. Suppose that G is a group and that H is a subgroup of G such that [G : H] = 2. Suppose that a, b G, but a H and b H. Prove that ab H. Solution: Since [G : H] = 2, it follows that H is a normal subgroup of G. Consider the quotient group G/H. It is a group of order 2. The identity element in that group is H. The other element in that group is x = ah since a H. However, this element of G/H is also bh since b H. That is, x = ah = bh. Since G/H has order 2, the element x is its own inverse. That is, xx = ahbh = H. Therefore, abh = H. Therefore, ab H, as stated. F. This is a question about the group S 5. It concerns the element ( 1 2 3 4 5 σ = 3 5 1 2 4 (a Write σ as a product of disjoint cycles. We have σ = (13(254.
(b Determine the order of σ. The order of σ is equal to 6, the least common multiple of the lengths of the cycles in the disjoint cycle decomposition for σ. (c Is σ A 5? We have σ = (13(254 = (13(24(25. Hence σ is an odd permutation. Hence σ A 5. (d Find an element τ S 5 such that τστ 1 = σ 1. We have σ 1 = (13 1 (254 1 = (13(245. Let τ S 5 be defined as follows: τ(1 = 1, τ(3 = 3, τ(2 = 2, τ(5 = 4, τ(4 = 5. Then, by the Conjugacy Principle, we have as we wanted. τστ 1 = (τ(1 τ(3 ( τ(2 τ(5 τ(4 = (13(245 = σ 1, (e True or False: Every element of S 5 which has the same order as σ is conjugate to σ in S 5. Justify your answer. The statement is true. The element σ has order 6. However, let σ be any element of order 6 in S 5. We can express σ as a product of disjoint cycles of various lengths. The sum of the lengths must be 5 and their least common multiple must be 6, the order of σ. Thus, σ must be a product of two disjoint cycles, one of length 2 and one of length 3. Therefore, σ has the same cycle decomposition type as σ. As explained in class, this implies that σ is a conjugate of σ in S 5.