Abstract Algebra, HW6 Solutions. Chapter 5
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1 Abstract Algebra, HW6 Solutions Chapter 5 6 We note that lcm(3,5)15 So, we need to come up with two disjoint cycles of lengths 3 and 5 The obvious choices are (13) and (45678) So if we consider the element (13)(45678), we get an element of order 15 Since we have (13)(45678) (1)(13)(45)(46)(47)(48), as the multiplication of six -cycles, (13)(45678) is an even permutation, consequently in A 8 1 Since any permutation of a finite set can be expressed as the multiplication of -cycles, we can express α as the multiplication of -cycles Let α β 1 β β n 1 β n for some positive integer n where β i is a -cycle for every i 1,, n Then we get α 1 (β 1 β β n 1 β n ) 1 βn 1 βn 1 1 β 1 β 1 1 β n β n 1 β β 1 since the inverse of every -cycle is itself, ie β 1 i β i for each i 1,, n Note that α 1 also has the same number n of -cycles with α Hence, if α is even, then α 1 is even, if α is odd, then α 1 is odd 18a) First, let us express α and β as the multiplication of disjoint cycles: α (1345)(678) and β (3847)(56) Then let us calculate αβ: αβ (1345)(678)(3847)(56) ( ), which is already a product of disjoint cycles b) In this part, we use the expression given in the proof of Theorem 54: (a 1 a a k 1 a k ) (a 1 a k )(a 1 a k 1 ) (a 1 a ) So we get α (1)(13)(14)(15)(67)(68), β (3)(8)(4)(7)(56), αβ (16)(13)(17)(15)(18)(14)(1) 36 Let us consider the element α (134) S 4 Note that the order of α is 4 Obviously, H α is a cyclic subgroup of order 4 in S 4 since α is 4 For the second part of the problem, let us consider the elements α (1) and β (34) in S 4 Let H {ε, α, β, αβ} We claim that H is a non-cyclic subgroup of order 4 in S 4 (Other variations are also correct with α and β 1
2 any two disjoint two cycles in S 4 Another correct answer is given by H {ε, (1)(34), (13)(4), (14)(3)}) First let us show that H is closed under multiplication: Note that we have α and β In other words, we obtain α α 1 and β β 1 Since α and β are disjoint cycles, we get αβ βα Therefore, multiplication of any finite number of α with any finite number of β in any combination is of the form α i β j where i, j are positive integers If both i and j are even, then we have i n and j m for some positive integers n and m So we get α i β j α n β m (α ) n (β ) m ε n ε m ε If i is even, j is odd, then we have i n and j m + 1 for some positive integers n and m So we get α i β j α n β m+1 (α ) n (β ) m β ε n ε m β β If i is odd, j is even, then we have i n + 1 and j m for some positive integer n and m So we obtain α i β j α n+1 β m α(α ) n (β ) m αε n ε m α If both i and j are both odd, then we have i n + 1 and j m + 1 for some positive integers n and m So we get α i β j α n+1 β m+1 α(α ) n (β ) m β αε n ε m β αεβ αβ Hence H is closed under multiplication Since we know that α, β, and α and β are disjoint, we get αβ lcm(, ), ie αβ Namely, we have (αβ) 1 αβ Since every element has an inverse in H, we conclude that H is a subgroup of S 4 Since every nonidentity element in H has order and H 4, H is not cyclic
3 Chapter 6 4 We follow the steps given on page 1 STEP 1: Let us define a candidate for isomorphism between G and H: φ : G H a + b [ a b b a ] STEP : We need to prove that φ is one-to-one Let g 1 a 1 + b 1 and g a +b be two elements of G Assume that φ(a1 +b 1 ) φ(a +b ) Then we get a1 b 1 a b If two matrices are equal to each other, their corresponding entries are equal In other words, we obtain a 1 a and b 1 b So we get g 1 g Hence φ is one-to-one STEP 3: We need to prove that φ is onto For any element h H, we need to find an element g in G such that φ(g) h Let h be a b b a If we look at the element g a + b in G we see that φ(g) h So φ is onto STEP 4: we need to prove that φ is operation-preserving Let g 1 a 1 +b 1 and g a + b be two elements of G Let us look at the following calculations: φ(g 1 + g ) φ((a 1 + a ) + (b 1 + b ) ) a1 + a b 1 + b b 1 + b a 1 + a a1 b 1 a b + φ(a 1 + b 1 ) + φ(a + b ) φ(g 1 ) + φ(g ) So φ is operation-preserving Hence G and H are isomorphic For the second part of the problem: G is closed under multiplication because for every element a 1 + b 1, a + b in G we have (a 1 + b 1 )(a + b ) a1 a + a 1 b + a b 1 + b1 b (a 1 a + b 1 b ) + (a 1 b + a b 1 ) G, where we also use that rational numbers are closed under addition and multiplication 3
4 H is closed under multiplication because for any two elements a1 b 1 a b, H, we have [ a1 b 1 ] [ a b ] a1 a + b 1 b (a 1 b + b 1 a ) H a b 1 + a 1 b b 1 b + a 1 a To answer the last part of the question, we look at the following calculation: φ((a 1 + b 1 )(a + b )) φ((a1 a + b 1 b ) + (a 1 b + a b 1 ) ) a1 a + b 1 b (a 1 b + b 1 a ) a b 1 + a 1 b b 1 b + a 1 a a1 b 1 a b So we see that φ preserves multiplication as well φ(a 1 + b 1 )φ(a + b ) 36 We follow the steps on page 1 for this problem also: STEP 1: Let us define a candidate for isomorphism between G and H: φ : G H n 3n STEP : We need to prove that φ is one-to-one Let n 1 and n be two elements of G Assume that φ(n 1 ) φ(n ), then we have 3n 1 3n or n 1 n which implies that n 1 n So φ is one-to-one STEP 3: We need to prove that φ is onto Let 3n be an element of H We need to find an element in G which is mapped to 3n by φ Obviously, if we consider n we see that φ(n) 3n So, φ is onto STEP 4: We need to prove that φ is operation-preserving Let n 1 and n be two elements of G, then we have φ(n 1 + n ) φ((n 1 + n )) 3(n 1 + n ) 3n 1 + 3n φ(n 1 ) + φ(n ) So φ is operation-preserving Hence G and H are isomorphic For the second part of the question, we need to carry out the following calculations: φ((n 1 )(n )) φ((n 1 n )) 3(n 1 n ) 6n 1 n, 4
5 φ(n 1 )φ(n ) (3n 1 )(3n ) 9n 1 n Note that we have 6n 1 n 9n 1 n only if n 1 0 or n 0 Since the equality φ((n 1 )(n )) φ(n 1 )φ(n ) is not true for every n 1 and n, φ does not preserve the multiplication 5
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