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Math 3300, Abstract Algebra I Solutions to Homework 4 Problems

Chapter 5. Problem 16. Associate an even permutation with the number +1 and an oddpermutation with the number 1. Draw an analogy between the result of multiplying twopermutations and the result of multiplying their corresponding numbers +1 or 1.

Solution: Since we define a permutation as even or odd based on its expression as the

product of 2-cycles, we see that the parity of a product of permutations depends on theparities of the permutations we are multiplying. In fact, if we have two permutations, and , which we have written as the products of a and b 2-cycles, respectively, then wecan write their product, as the product of a + b 2-cycles. Hence, the product of twoeven permutations is an even permutation, of two odds will be even, and of an even and anodd will be odd. Analogously, multiplying +1 with itself or 1 with itself yields +1, whilemultiplying +1 and 1 (in either order) yields 1.

+1 1+1 +1 11 1 +1

even oddeven even odd

odd odd even

Chapter 5. Problem 18. Let =

1 2 3 4 5 6 7 82 3 4 5 1 7 8 6

and =

1 2 3 4 5 6 7 81 3 8 7 6 5 2 4

Write , and asa. products of disjoint cycles,b. products of 2-cycles.

Solution: Notice that =

1 2 3 4 5 6 7 82 4 6 8 7 1 3 5

. Thus,

a. = (12345)(678), = (23847)(56), and = (12485736).b. = (43)(42)(41)(45)(76)(78), = (48)(43)(42)(47)(56), and = (37)(35)(38)(34)(32)(31)(36)

Chapter 5. Problem 22. Let and belong to Sn. Prove that 11 is an even

permutation.

Solution: Since we know that can be written as a product of 2-cycles, we can easily defineits inverse as the product of the same 2-cycles, composed in the opposite order. In symbols:

if = 12 s then 1 can be written 1 = ss1 1. Since both and 1 can bewritten as the product ofs 2-cycles, they are either both even or both odd (By Theorem 5.5and the definition). Hence, if s is even, then both and 1 must be even. Likewise, if sis odd, then so are and 1. Similarly for , and 1 must either both be even or bothbe odd. We see that the product 11 is either the product of 4 even permutations, 4odd permutations, or 2 even and two odd permutations. Using the idea from Problem 16,we see that in any of these cases, we end up with an even permutation.

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Math 3300, Abstract Algebra I Solutions to Homework 4 Problems

Chapter 5. Problem 36. In S4, find a cyclic subgroup of order 4 and a non-cyclic subgroupof order 4.

Solution: Let = (1234). Then 2 = (13)(24), 3 = (1432), and 4 = . Hence, ={,,2, 3} is a cyclic subgroup of order 4.

Now, let = (12) and = (34). Then =

1, =

1, = (12)(34) and()1 = 11 = = , since and are disjoint 2-cycles. We can verify thatthe set {,,,} is a non-cyclic subgroup by examining its Cayley Table, shown below.

Chapter 5. Problem 46. Show that for n 3, Z(Sn) = {}.

Solution 1: Suppose Z(Sn+1), and let be some permutation in Sn. Then we canthink of in Sn+1 as the permutation which does the same thing to the first n letters, andfixes the last one. So, = must hold Sn.

Let = (123 . . . n), and suppose sends (n + 1) to k = n + 1 (well write this as(n + 1) = k). Then sends (n + 1) to k + 1 ((n + 1) = k + 1), and sends (n + 1)to k, since fixes (n + 1) ((n + 1) = (n + 1) = k), which must be equal to k + 1, since = since Z(Sn+1). But k = k + 1, hence must actually fix (n + 1), that is(n + 1) = n + 1. Since was arbitrarily chosen from Z(Sn+1), we see that, in fact, everyelement of Z(Sn+1) must fix (n + 1). This, and our initial observation that these elementsmust commute with each permutation in Sn shows that, in fact, Z(Sn+1) Z(Sn).

This creates a chain of containment, and we know that Z(S3) Z(Sn) for any n 3.But upon examination, we see that the only element in Z(S3) is the identity, , which weknow to be contained in the center of every group. Hence, Z(Sn) = {} n 3.

Solution 2: Suppose Sn is not the identity element. We claim that Z(Sn) = { Sn : = Sn}. To prove this, we need only find one Sn for which = .As is not the identity, there is a number a such that (a) = a. Set b = (a). Since n 3,there is a distinct number c such that c = a and c = b. Set = (ac), the 2-cycle thatexchanges a and c and fixes everything else (including b). Now consider

()(a) = ((a)) = (c)

( )(a) = ((a)) = (b) = b = (a).Since Sn, is one-to-one, so if = , then we must have

(c) = ()(a) = ( )(a) = (a),

so that c = a. This is a contradiction, since we chose c such that c = a. Thus (c) = (a),so we are forced to conclude that = . Hence, given = (where denotes the identitypermutation in Sn), there exists a 2-cycle Sn such that = . Therefore, we concludethat Z(Sn), so Z(Sn) = is the trivial subgroup of Sn.