## Math 417 Homework 7: Due Friday March 14

Instructions: You can discuss these problems with others, but write up your solutions on your own (i.e., don't just copy someone else's solutions, else the feedback I give you won't help you much). Please be neat and write in full sentences.
Do four of the five problems.
• [1]
• Determine the number of elements in S5 of order 3.

An element x of order 3 must be a product of disjoint 3-cycles in order for the least common multiple of the lengths of its cycles to be 3. But since x is in S5, this means the number of disjoint 3 cycles is 1; i.e., x is itself a 3-cycle. Thus x = (a b c), and there are 5 choices for the a, 4 for the b and 3 for the c, giving 5(4)(3) = 60 ways to write down such an x. Since (a b c) = (b c a) = (c a b), we must divide by 3 to get a count of the number of different such x. Thus there are 20 elements in S5 of order 3.

• Determine the number of elements in S6 of order 3.

Again, an element x of order 3 must be a product of disjoint 3-cycles in order for the least common multiple of the lengths of its cycles to be 3. But since x is now in S6, this means the number of disjoint 3 cycles can be either 1 or 2; i.e., x is either itself a 3-cycle (a b c) or x is a product of two disjoint 3-cycles, (a b c)(d e f). There are 6(5)(4)/3 = 40 3-cycles in S6. To count the elements of S6 of the form (a b c)(d e f), note that a, b c, d, e, and f comprise all of the numbers 1 to 6. Since we can rotate the entries of a cycle and since we can put either cycle first (since being disjoint they commute), we can assume that a = 1, and we can assume that d < e and d < f. Thus d is determined once b and c are chosen, so to fill in for b, c, e and f we have only 5(4)(2)(1) = 40 different results. Thus there are 80 different elements of order 3 in S6.

• [2] Problem #46 on p. 113: Prove that Z(Sn) = { e } if n > 2.

Clearly, e is in Z(Sn). To show that no other permutation is in Z(Sn), for each x in Z(Sn) we have to find a permutation y such that xy is not equal to yx. So say x is a 2-cycle, (a b). Since n > 2, we can take y = (a c), where c is not a or b. Since xy = (a c b), while yx = (a b c), we see xy and yx are different, so x is not in Z(Sn). If x is a product of 2 or more disjoint 2-cycles, this product involves, say (a b) and (c d), where a, b, c and d are all different. Let y = (a c). Then xy takes a to d, but yx takes a to b, so xy and yx are again different. Now say x, as a product of dijoint cycles, involves and r-cycle, with r > 2. We may assume this r-cycle is (a b c ...); i.e., its forst three entries are a, b and c. Let y = (a b). Then xy takes b to b, but yx takes b to c, so we are done.

• [3] Problem #2 on p. 129: Find Aut(Z).

We know the identity, idZ, is in Aut(Z). Say f is also in Aut(Z), but not the identity. Let n = f(1). Then f(Z) = { f(i) : i is an integer } = {if(1) : i is an integer } = {in: i is an integer } = the set of all integer multiples of n. But 1 is not an integer multiple of n unless n = 1 or n = -1. If n = 1, then f(i) = 1i = i for all i, hence f = idZ, contrary to assumption. Thus n must be -1, hence f(i) = -i for all i. This is bijective and f(i + j) = -(i + j) = -i + -j = f(i) + f(j), so this also gives an isomorphism. Thus Aut(Z) has two elements, idZ and f. (Note that ff = idZ, so sedning idZ to 1 and f to -1 establishes an isomorphism from Aut(Z) to the group {1, -1}, where the group operation of {1, -1} is ordinary multiplication.)

• [4] Problem #10 on p. 130: Let G be a group. Prove that the mapping a(g) = g-1 for all g in G is an isomorphism if and only if G is abelian.

=>: Assume a(g) = g-1 is an isomorphism. Then for any elements x and y in G we have: yx = (y-1)-1(x-1)-1 = (x-1y-1)-1 = a(x-1y-1) = a(x-1)a(y-1) = (x-1)-1(y-1)-1 = xy, hence x and y commute, so G is abelian.

<=: Assume that G is abelian. Then for any x and y in G we have: a(xy) (xy)-1 = (yx)-1 = x-1y-1 = a(x)a(y). Since a(a(x)) = ((x)-1)-1 = x, we see that aa = idG, so a is both injective and surjective, hence bijective. Thus a is an isomorphism.

• [5] Problem #22 on p. 130: U(20) is not isomorphic to U(24).

Although |U(20| = phi(20) = phi(4)phi(5) = (4 - 2)(5 - 1) = 8 = (3 - 1)(8 - 4) = phi(3)phi(8) = phi(24) = |U(24)|, every nontrivial element of U(24) has order 2, whereas 3 as an element of U(20) has order 4. Thus by Theorem 6.2(4), there can be no isomorphism between U(20) and U(24).