Exam 2 solutions

Math 116: Finashin, Pamuk, Pierce, Solak Tuesday, May 3, 2011

Instructions: Please, be accurate and show clearly the logic of your solutions. Only the answers are not enough: indicate your calculations and arguments. Problem 1. (9 pts) Of the set {1, 2, 3, 4, 5, 6, 7, 9}, let σ ∈ S_{9} be the permutation

[\begin{matrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 3 & 7 & 4 & 1 & 6 & 8 & 2 & 9 & 5 \end{matrix}] .

(a) Write σ as a product of disjoint cycles. Solution. σ = (134)(27)(5689).

(b) Find σ^{100}. (Hint: find the order of σ).

Solution. The length of σ is the least common multiple of the length of the disjoint cycles. Hence |σ| = lcm(3, 2, 4) = 12. Thus

σ^{100} = σ^{96}σ^{4} = σ^{4} = (134).

(c) Determine whether permutation σ is even or odd. Solution. Since we can write σ = (14)(13)(27)(59)(58)(56) as a product of even number of transpositions, σ is an even permutation.

Problem 2. (6 pts) Consider transpositions α = (12), β = (23), γ = (34) in S_{4}.

(a) Find the product αβγ.

Solution. αβγ = (12)(23)(34) = (1234).

(b) Find g ∈ S_{4} such that β = gαg^{−1}.

Solution. Note that g is not unique. Since g(1) = 2 and g(2) = 3 we may write g as

g = (12)(23) = (123)

.

Problem 3. (8 pts)

(a) Find the number of cosets for subgroups 3ℤ ⊂ ℤ and ℤ ⊂ ℚ. In the first case, give an explicit list of these cosets.

Solution. The cosets of 3ℤ in ℤ are 3ℤ, 1 + 3ℤ, and 2 + 3ℤ: there are three of them.

ℤ has infinitely many cosets in ℚ [since if a and b are distinct elements of ℚ between 0 and 1, then a + ℤ and b + ℤ are distinct cosets].

(b) Find the group multiplication table of ℤ/3ℤ (pay attention that the group operation is addition).

Solution. \begin{aligned}& 3ℤ & 1 + 3ℤ & 2 + 3ℤ \\ 3ℤ & 3ℤ & 1 + 3ℤ & 2 + 3ℤ \\ 1 + 3ℤ & 1 + 3ℤ & 2 + 3ℤ & 3ℤ \\ 2 + 3ℤ & 2 + 3ℤ & 3ℤ & 1 + 3ℤ \end{aligned}

(c) Prove that every element of ℚ/ℤ has finite order.

Solution. Every element of ℚ/ℤ is k/n+ℤ for some k and n in ℤ such that n > 0. Then n(k/n + ℤ) = k + ℤ = ℤ, the identity in ℚ/Z. Thus the order of k/n + ℤ is less than or equal to n: in particular, it is finite.

Problem 4. (7 pts) Assume that G is a group of prime order p.

(a) What can be the order of an element g ∈ G ? (Explain !)

Solution. By the Lagrange Theorem, the order of an element divides the order of a group. In particular, the order of an element of G divides p; so the order is 0 or 1.

(b) Prove that G is a cyclic group.

Solution. The only element of G of order 1 is the identity, e. If a ∈ G ∖ {e}, then the order of a is p, by part (a). So the cyclic subgroup ⟨a⟩ of G has the same order as G itself; so that cyclic subgroup is G.

Problem 5. (10 pts) Suppose that H and N are subgroups of a group G. Assume that subgroup N is normal, and let K = H ∩ N.

(a) Prove that K is a normal subgroup of H.

Solution. Since H, N are subgroups of G then e ∈ H and e ∈ N, so e ∈ H∩N = K i.e. K is non-empty. Let a, b ∈ K consider ab^{−1}, since a, b ∈ H and a, b ∈ N and they are subgroups of G then ab^{−1} ∈ H and ab^{−1} ∈ N, so ab^{−1} ∈ K. Therefore, K is a subgroup of H. Let h ∈ H and a ∈ K, then a ∈ H and a ∈ N, so hah^{−1} ∈ H and since N is normal hah^{−1} = n for some n ∈ N which means hah^{−1} ∈ N, therefore hah^{−1} ∈ H ∩ N = K. Hence, K is a normal subgroup of H.(OR: ha = bh for some b ∈ N since N is normal, but b = hah^{−1} which is in H as well, so b ∈ K, hence hK = Kh for all h ∈ H.)

(b) Let A = {x ∈ G: xK = Kx}. Prove that H ⊆ A.

Solution. By part (a)K is normal in H, so we know that hK = Kh for all h ∈ H, but then h ∈ A for all h ∈ H. Hence H ⊆ A.

(c) Prove that A is a subgroup of G.

Solution. (i) Since eK = K = Ke, we have e ∈ A. (ii) If x ∈ A, so that xK = Kx, then K = x^{−1}xK = x^{−1}Kx, so Kx^{−1} =

x^{−1}Kxx^{−1} = x^{−1}K, and so x^{−1} ∈ A. (iii) If x, y ∈ A, then xK = Kx and yK = Ky, so xyK = xKy = Kxy, hence

xy ∈ A.

Problem 6. (5 pts) Let H and K be normal subgroups of G such that G/H has order 5, and G/K has order 3. Prove that for any g ∈ G, g^{15} ∈ K ∩ H.

Solution. Let g ∈ G, consider the element gH ∈ G/H. Since G/H has order 5 then by Lagrange Thm. o(gH)|5 by problem 4(a) o(gH) = 5 or o(gH) = 1. If o(gH) = 1 then gH = eH = H ⇒ g ∈ H ⇒ g^{5} ∈ H. If o(gH) = 5 then since H is a normal subgroup of G

eH = (gH)^{5} = gHgHgHgHgH = gHgHgHggH = ⋯ = g^{5}H

which means g^{5} ∈ H.

Similarly, o(gK)|3 then o(gK) = 3 or o(gK) = 1. If o(gK) = 1 then g^{3} ∈ K. If o(gK) = 3 then since K is a normal subgroup of G eK = (gK)^{3} = g^{3}K which means g^{3} ∈ K.

Now, since g^{1}5 = (g^{5})^{3} ∈ H and g^{1}5 = (g^{3})^{5} ∈ K then g^{1}5 ∈ K ∩ H.

Problem 7. (6 pts) Consider subsets 4ℤ ⊂ ℤ, {0} ⊂ ℤ, ∅ ⊂ ℤ, ℤ ⊂ ℚ, (0, ∞) ⊂ ℝ, {[0], [1], [2], [4]} ⊂ ℤ_{6}, M_{2}(ℤ) ⊂ M_{2}(ℝ).

(a) Which of the above examples are NOT subrings? Why they are not subrings ?

Solution. ∅ ⊂ ℤ, because a subring should be non-empty,

(0, ∞) ⊂ ℝ, because for number x ∈ (0, ∞) its opposite, −x, does not belong to (0, ∞),

{[0], [1], [2], [4]} ⊂ ℤ_{6}, because [1] + [2] = [3] ∉ {[0], [1], [2], [4]}.

(b) Which ones are NOT ideals? Why not ?

Solution. An ideal is a subring, thus, the three examples in part (a) which are not rings are also not ideals. There are two more examples:

ℤ ⊂ ℚ (the product of a rational and an integers may be NOT an integer), and

M_{2}(ℤ) ⊂ M_{2}(ℝ) (the product of an integer matrix with a real matrix may be NOT an integer matrix).

(So, the only examples of ideals here are 4ℤ ⊂ ℤ, and {0} ⊂ ℤ.)

Problem 8. (9 pts) In each of the following two subrings of ℤ_{12}, find a unity or show that there is no unity (multiplicative identity).

(a) 2ℤ_{12} = {[0], [2], [4], [6], [8], [10]}

Solution. If [2n] is the unity, then [2][2n] = [2], that is 4n = 2 mod 12. But 4n − 2 is not divisible by 4, and thus, by 12. So, there is no unity in 2ℤ_{12}.

(b) 3ℤ_{12} = {[0], [3], [6], [9]}

Solution. [9] is the unity in 2ℤ_{12}, because [9][0] = [0], [9][3] = [27] = [3], [9][6] = [54] = [6], [9][9] = [81] = [9].

(c) Find all the zero divisors in the ring ℤ_{12}.

Solution. The zero divisors are [2], [3], [4], [6], [8], [9], [10], because in ℤ_{12}, we have

[2][6] = [3][4] = [8][9] = [10][6] = [0].