Homework 12 sol PDF

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Homework 12 SolutionSections 7, 8. deadline: Dec 13, 1:00 pm Do not abbreviate your answer. Write everything in full sentences. Write your answer neatly. If I couldn’t understand it, you’ll get 0 point. You may discuss with your classmates. But do not copy directly. Let α=[1 2 3 4 5 6 7 83 4 1 2 6 7...

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MATH 3005 Homework

Han-Bom Moon

Homework 12 Solution Sections 7.5, 8.1. deadline: Dec 13, 1:00 pm • Do not abbreviate your answer. Write everything in full sentences. • Write your answer neatly. If I couldn’t understand it, you’ll get 0 point. • You may discuss with your classmates. But do not copy directly.

1. Let α=

"

1 2 3 4 5 6 7 8 3 4 1 2 6 7 5 8

#

,

β=

(a) Compute αβ, βα, and α−1 . " # 1 2 3 4 5 6 7 8 αβ = , 7 6 2 3 1 4 8 5 α−1 =

"

"

1 2 3 4 5 6 7 8 6 5 4 1 3 2 8 7

βα =

"

1 2 3 4 5 6 7 8 3 4 1 2 7 5 6 8

β = (162534)(78),

#

.

αβ = (1785)(2643).

(c) Write α, β, αβ as products of 2-cycles. α = (13)(24)(57)(56),

β = (14)(13)(15)(12)(16)(78),

αβ = (15)(18)(17)(23)(24)(26). (d) Find |α| and |β|. |α| = lcm(2, 2, 3) = 6,

|β| = lcm(6, 2) = 6.

2. Let α = (1357)(462),

β = (265)(13)(47).

(a) Write αβ, βα, α−1 as products of disjoint cycles. αβ = (154)(67),

βα = (273)(45),

1

.

1 2 3 4 5 6 7 8 4 1 6 5 2 8 3 7

(b) Write α, β, αβ as products of disjoint cycles. α = (13)(24)(567),

#

α−1 = (1753)(264).

#

,

MATH 3005 Homework

Han-Bom Moon

(b) Write α, β, αβ as products of 2-cycles. α = (17)(15)(13)(42)(46),

β = (25)(26)(13)(47),

(c) Write α, β, αβ as arrays. " # 1 2 3 4 5 6 7 α= , 3 4 5 6 7 2 1 αβ =

"

β=

"

αβ = (14)(15)(67).

1 2 3 4 5 6 7 3 6 1 7 2 5 4

1 2 3 4 5 6 7 5 2 3 1 4 7 6

#

#

,

.

(d) Find |α| and |β|. |α| = lcm(4, 3) = 12,

, |β| = lcm(3, 2, 2) = 6.

3. Find the maximum of |α| for α ∈ S12. We can obtain an element with maximum order if α is a product of disjoint cycles α1 , α2 , · · · , αk such that the least common multiple of their lengths is the largest one. By considering all possible cases, one can check that the case that α is a product of three disjoint cycles of length 3, 4, and 5 gives the largest order. In this case, the order is 60. 4. Let G be a subgroup of Sn . (a) If there is an odd permutation α ∈ G, then show that exactly half of the elements of G are even. (How can we show the fact that the given set has an even number of elements?) Let E be the set of even permutations in G, and let O be the set of odd permutations in G. Define a map f : E → O, as f (β) = αβ. This map is welldefined, because a product of an odd permutation and an even permutation is odd. f is one-to-one, because f (β1 ) = f (β2 ) ⇒ αβ1 = αβ2 ⇒ β1 = β2 from the cancellation of α. The map f is onto because for every odd permutation γ ∈ O, α−1 γ ∈ E and f (α−1 γ) = αα−1 γ = γ. So f is a bijection and |E| = |O|. So we have |G| = |E| + |O| = 2|E| and |G| is even. (b) Suppose that G has odd number of elements. Show that G ≤ An . (Hint: Use (a).) If G has an odd permutation, then |G| must be even by (a). So if |G| is odd, then all elements of G are even permutations. Therefore G ≤ An . 5. Find all right cosets of K in G. 2

MATH 3005 Homework

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(a) G = Z, K = h4i. By division algorithm, every integer can be written as 4m+r for r = 0, 1, 2, 3. Because 4m ∈ h4i = K, K + 4m + r = K + r. Therefore there are only four cosets, K, K + 1, K + 2, and K + 3. (b) G = Z∗9 , K = h8i. G = {1, 2, 4, 5, 7, 8} and K = h8i = {1, 8}. So there are |G|/|K| = 3 cosets. K = {1, 8}, K2 = {2, 7}, and K4 = {4, 5} are those three cosets. (c) G = D4 , K = hhi. Because |G| = |D4 | = 8 and |K| = |h| = 2, there are four distinct cosets. K = {e, h}, Kr1 = {r1 , d }, Kr2 = {r2 , v }, Kr3 = {r3 , t}. (d) G = A4 , K = {e, (12)(34), (13)(24), (14)(23)}. Because |G| = |A4 | = 4!/2 = 12 and |K| = 4, there are three cosets. K = {e, (12)(34), (13)(24), (14)(23)}. K(123) = {(123), (243), (142), (134)}, and K(132) = {(132), (143), (234), (124)}. 6. Let G = GL2 (R) and H = SL2 (R). For A ∈ GL2 (R), show that H A = {B ∈ GL2 (R) | det B = det A}. Let X = {B ∈ GL2 (R) | det B = det A}. If B ∈ H A, then there exists C ∈ H such that B = CA. Then det B = det(CA) = (det C)(det A) = det A because C ∈ H = SL2 (R). So B ∈ X and H A ⊂ X. Conversely, if B ∈ X, then det(BA−1 ) = det B det(A−1 ) = det B(det A)−1 = (det A)(det A)−1 = 1 and BA−1 ∈ H. Therefore B = BA−1 A ∈ H A and X ⊂ H A. Thus X = H A. 7. Let G be a group such that |G| = pn for some prime number p and a positive integer n. Show that G has an element of order p. Take a ∈ G such that a 6= e. By a corollary of Lagrange’s theorem, |a| | |G| = pn . k−1 k−1 So |a| = pk for some 1 ≤ k ≤ n. Then |ap | = pk /pk−1 = p. So ap is what we want. 8. Let G be a group of order 35. (a) If H and K are two distinct subgroups of order 7, show that H ∩ K = {e}. Suppose that H and K are two distinct subgroups of order 7. Then H ∩ K ≤ K . So |H ∩ K |||K | = 7. Thus |H ∩ K | = 1 or 7. If |H ∩ K | = 7, then H = K. Thus |H ∩ K| = 1. This implies H ∩ K = {e}. 3

MATH 3005 Homework

Han-Bom Moon

(b) Prove that G has an element of order 5. (Hint: By using (a), show that the number of elements of order 7 is a multiple of 6.) If G is cyclic and G = hai, then |a7 | = 35/7 = 5. Now suppose that G is not cyclic, then every element has the order 1, 5, or 7. Also the only one element of order 1 is e. For two nonidentity elements a, b ∈ G of order 7, hai ∩ hbi ≤ hai so |hai ∩ hbi| | |hai| = 7. Therefore if hai 6= hbi, |hai ∩ hbi| = 1 and hai ∩ hbi = {e}. Furthermore, in hai, except e, all other elements have order 7 so there are 6 elements of order 7 in hai. In summary, the number of order 7 elements in G is 6k for some k ∈ N. Now |G − {e}| = 34 is not the multiple of 6. Therefore there must be an element of order 5. (c) Show that G has an element of order 7. If G is cyclic and G = hai, then |a5 | = 35/5 = 7. If G is not cyclic, then by the same argument in the previous problem, we can see that the number of order 5 elements in G is 4k for some k ∈ N. But |G − {e}| = 34 is not a multiple of 4. Therefore there must be an element of order 7. (d) Furthermore, assume that G is abelian. Show that G is cyclic. (Hint: Let α, β ∈ G be elements such that |α| = 5 and |β| = 7. What is |αβ|?) By (b) and (c), there are two element a, b ∈ G such that |α| = 5, |β| = 7. We claim that |αβ| = 35. Note that from hαi ∩ hβi ≤ hαi, |hαi ∩ hβi| | |hαi| = 5. Similarly, we have |hαi∩hβi| | |hβi| = 7. So |hαi∩hβi| = 1 and hαi∩hβi = {e}. If |αβ| = n, then αn β n = (αβ)n = e. So α−n = β n ∈ hαi ∩ hβi = {e}. Thus β n = e and 7 = |β| | n. And αn = (α−n )−1 = e−1 = e, so 5 = |α| | n. Therefore n ≥ lcm(7, 5) = 35. The maximum value of an order of αβ ∈ G is 35. So |αβ| = 35 and G = hαβi.

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