Hints for Homework 4 Chapter 2: 14 This is the problem I reassigned. Here's a mistake that some people made last time: you can't choose just any a, b, and c and have that ab = ca. The group has the property that for any a, b, c such that ab = ca then b = c. You want to prove that the group is abelian. Your goal is to show that xy = yx for any group elements x, y. Match this equation up to b and c in the special property and find an element a that makes the property applicable. 17 Gallian suggests to use Exercise #16. That seems like a good way to start this. In that exercise, we have that (ab)^-1 = b^-1 a^-1, so apply that to the statement in #17 and go from there. 36 Look at theorem 4.4 and its corollary on page 80 and try to apply them. Chapter 3: 11 When Gallian says find "a" group such that certain properties hold, he doesn't mean the same group in each case. This wording may be confusing, but his meaning is clear from context. The group element ab has a well-defined order, which is a positive integer. The order can't simultaneously be 3, 4, and 5. Note also that the order of element ab doesn't have to be the same number for all choices of a and b. For each case, find a group and two particular elements of that group with order 2 and the property that |ab| = 3, 4, or 5, respectively. 14 Use the two-step subgroup test to show that the intersection is closed under composition and inversion. That is, show that for any two elements a, b that are in both H and K, that ab is in the intersection and that a^-1 is in the intersection. (Give a clear written explanation for this proof. For the question "Can you see...", just give a brief answer. Please think about it, but "no" is a valid response.) 17 and 18 Let the elements of the group be indexed as { a_1 = e, a_2,...,a_n }. The row i column j entry (i,j) of the Cayley matrix is the element a_i a_j . Given two elements a_i, a_j, if a_j is in the centralizer of a_i, that is, in C(a_i), then a_i a_j = a_j a_i. This means that the (i,j) entry is equal to the (j,i) entry. 24 Write out the conditions for an element to be in C(a) and C(a^3). Show that they are equal by either (1) writing a reversible argument that shows x is in C(a) iff x is in C(a^3), or (2) show that C(a) is contained in C(a^3) and vice versa. One direction is easy. The other direction is harder. Start with x a^3 = a^3 x and multiply on both the left and right by a power of a that lets you cancel out a^5 and show that x commutes with a different power of a. Then go from there. (There are probably a couple of ways to do this problem.) Chapter 4: 7 (contributed by Karen Reed) For #7 you should answer two questions. 1. Why is the group not cyclic? Look at U(8). Determine the orders of the elements. Does U(8) = for any a in U(8)? 2. Look at the subgroups of U(8). Hint: Any two elements of {3, 5, 7} will generate U(8). Ex. <3,5> 3*5 =15 3*5 = 7 mod8 16 Consider the factorization of 120 into primes as 2^3 * 3 * 5. Apply the Fundamental Theorem of Cyclic Groups. The orders of the groups in the chain must have must successively divide each other (a_{i+1} divides a_i). If the quotient of one order by the next has more than one prime factor, then you're missing an intermediate step. 39 (contributed by Michelle Yakaboski) Based on theorem 4.3 the fundamental theorem of cyclic groups, the group, G, has as many subgroups as the number of divisors of the order of the group. The number 32 has 6 divisors, that will work! 32 has the divisors, 1, 2, 4, 8, 16, 32. Therefore: |a|=32, < a > = G |a^2|=16, < a^2 > = {1, a^2, a^3...a^30} |a^4|=8, < a^4 > = {1, a^4, a^8...a^28} |a^8|=4, < a^8 > = {1, a^8, a^16, a^24} |a^16|=2, < a^16 > = {1, a^16} < a^32 > = < e > A group with exactly 6 subgroups is < a^32 >, with the subgroups: < a >, < a^2 >, < a^4 >, < a^8 >, < a^16 >, < e >.