Math 31 Show Your Work! Page 1 of (10) Complete the following definitions.

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1 Math 1 Show Your Work! Page 1 of 4 1. (10) Complete the following definitions. (a) A nonempty set G together with a binary operation (a,b) ab from G G to G is called a group if... ANS: if (1) the operation is associative: (ab)c = a(bc) for all a,b,c G, (2) there is an identity element e G such that ae = a = ea for all a G, and () given a G there is a b G such that ab = e = ba. (b) If G is a group and a G, then the order of a is... ANS: the least positive integer n such that a n = e. If there is no such integer, then we say that the order of a is infinite. (c) If H is a subgroup of a group G, then the left coset of H in G containing a is... ANS: ah := {ah : h H }. (d) A permutation σ S n is called even if... ANS: we can write σ as the product of an even number of 2-cycles. 2. (10) Suppose that σ S 8 is the permutation given by σ = (a) Write σ as the product of disjoint cycles. ANS: We have σ = (17)(2648). (I accepted this without any justification.) (b) Is σ an odd or even permutation? ANS: Since (17)(2648) = (17)(1)(28)(24)(26), σ is odd. (c) What is the order of σ in S 8? ANS: By Ruffini s Theorem, σ = lcm(, 4) = 12. ( ) (10) Let Z 12 be the cyclic group { 0, 1, 2,, 4, 5, 6, 7, 8, 9, 10, 11 } with the operator of addition modulo 12.

2 Math 1 Show Your Work! Page 2 of 4 (a) List all the generators of Z 12. ANS: By a Theorem proved in lecture, the generators are the elements of U(12) = {1,5,7,10 }. (That is, the generators are those positive integers relatively prime to 12.) (COMMENT: Make sure you answer the question asked. A surprising number of people answered 1, 5, 7, 10 which is not a list of generators, but four different ways to write Z 12.) (b) List all the subgroups of Z 12. ANS: There is one subgroup for each divisor d of 12, and that subgroup will have order 12/d. Since the divisors are 1, 2,, 4, 6, 12, the subgroups are, respectively: 1, 2,, 4, 6, 0. (COMMENT: I didn t take off if you missed out 1 = Z 12, but Z 12 is a subgroup of itself.) (c) Draw the lattice diagram for the subgroups of Z 10. ANS: (5) What is the order of a = 21 in Z 120? ANS: By a Theorem proved in lecture, 21 = 120 gcd(120,21) = 120 = (5) State Lagrange s Theorem. ANS: If H is a subgroup of a finite group G, then the order of H divides the order of G. Moreover the number of distinct left (or right) cosets of H in G is G / H. 6. (10) Recall that if a is an element in a group G, then C(a) is the centralizer of a in G. Show that if g G, then C(a) and C(gag 1 ) are isomorphic. ANS: (COMMENT: this one proved to be more difficult than I anticipated. Please read the solution carefully whether or not you received a good score. EVERYONE should have been trying to craft a solution using the steps we outlined in lecture.)

3 Math 1 Show Your Work! Page of 4 Step one: we need a map φ : C(a) C(gag 1 ). A good guess is to let φ(x) := gxg 1. BUT, we need to see that if x C(a), then φ(a) C(gag 1 ). But if x C(a), then which, since x C(a) is φ(x)gag 1 = (gxg 1 )(gag 1 ) = gxag 1 = gaxg 1 = (gag 1 )(gxg 1 ) = (gag 1 )φ(x). Hence φ(x) C(gag 1 and φ is a map of C(a) into C(gag 1 ). Step two: Show φ is one-to-one. This is easy, if φ(x) = φ(y), then gxg 1 = gyg 1, and cancellation implies that x = y. This shows that φ is one-to-one. Step three: Show that φ is onto. Let y C(gag 1 ). We first need to see that g 1 yg C(a). But which, since y C(gag 1 ), is (g 1 yg)a = g 1 y(gag 1 )g = g 1 (gag 1 )yg = a(g 1 yg). That is, g 1 yg C(a) and then we can notice that φ(g 1 yg) = y, so that φ is onto. Step four: We still have to see that φ is multiplicative. But φ(xy) = gxyg 1 = gxg 1 gyg 1 = φ(x)φ(y). This completes the proof that C(a) and C(gag 1 ) are isomorphic. 7. (10) We say that a group satisfies middle cancellation if axb = cxd implies ab = cd. Show that a group that satisfies middle cancellation must be Abelian. ANS: (COMMENT: I am actually sorry I chose this one. It caused a lot of trouble, and I didn t mean it to be hard. Remember to show a group is Abelian, you have to show that ab = ba for all a,b G.) Let a,b G. Let x = a 1, c = b and d = a. Then axb = aa 1 b = b and cxd = ba 1 a = b. Then middle cancellation says that ab = ba. Since a and b were arbitrary elements of G, G must be Abelian. 8. (8) What are the orders of elements of A 5, and how many elements of each order are there? ANS: As we did in lecture (and as in the text), we let (n) denote a generic cycle of length n. Since a cycle of odd length is an even permutation, the possible disjoint cycle decompositions of elements of A 5 are (5), ()(1)(1), (2)(2)(1), and (1)(1)(1)(1)(1). Since these have orders, 5,, 2 and 1, respectively, these represent the possible orders of elements in A 5.

4 Math 1 This page is for scratch work Page 4 of 4 and 5 4 Next, we notice that there are 5! 5 = 24 elements of order 5 in A 5, 5 4 = 20 elements of order, = 15 elements of order 2 and one element of order (7) Suppose that p is a prime and the G has order p n. Prove that the center of G cannot have order p n 1. (Hint: consider the centralizer of an element not in the center.) ANS: (COMMENT: I was pleased to see that many people did a good job on this one. Still, I hope everyone will compare their solution to mine. Also, we ll see a better way to prove this in class Friday.) We ll do a proof by contradiction. Suppose that Z(G) = p n 1. Then there is an a G such that a / Z(G). (Alternatively, we could write a G \ Z(G).) But clearly, Z(G) C(a). Since a C(a), we must have C(a) > Z(G) = p n 1. By Lagrange s Theorem, C(a) must divide G = p n, so we must have C(a) = p n. But then C(a) = G. However, this implies that a Z(G), which contradicts our choice of a. This proves that we can t have Z(G) = p n 1.

5 NAME : Math 1 21 July 2010 Small Midterm Instructions: Complete all 9 problems. Be sure to carefully and completely justify your assertions. Other than asking your instructor for clarification, no calculators, notes, books or outside help of any kind is permitted. Problem Points Score Total 75