Lecture8 Copy

Lecture8 Copy

Warmup Recall, a group H is cyclic if H can be generated by a single element. In other words, there is some element x H for which P Multiplicative notation: H x` ` Z x , “t | P u“x y Additive notation: H `x ` Z x . “t | P u“x y You try: What are the cyclic subgroups of S4? Last time: Cyclic groups A group H is cyclic if H can be generated by a single element. In other words, there is some element x H for which P mult: H x` ` Z x add: H `x ` Z x . “t | P u“x y “t | P u“x y Proposition If H x ,then H x .Namely, “x y | |“| | (1) H n i↵ xn 1 and 1,x,x2,...,xn 1 are all distinct, | |“ “ ´ (2) H i↵ xa xb for all a b. | |“8 ‰ ‰ Theorem Any two cyclic subgroups of the same order are isomorphic. In particular, if x y , then are both cyclic groups of |x y| “ |x y| † 8 order n,then x y defined by xk yk x yÑx y fiÑ is an isomorphism; and if x ,then |x y| “ 8 Z x defined by k xk Ñx y fiÑ is an isomorphism. Notation: Let Zn be the cyclic group of order n (under multiplication). How many generators? If H x ,(read“H is generated by x, with or without relations), “x y we call x a generator, though it generally is not the only generating element. 3 Example: In D8, r r . x y“x y 1 1 ` Example: If H x ,thenH x´ x´ ` Z . “x y “x y“tp q | P u Proposition Let H x . “x y 1. Assume x .ThenH xa i↵ a 1. | |“8 “x y “˘ 2. Assume x n.ThenH xa i↵ a, n 1. | |“ “x y p q“ In particular, the number of generators of H is φ n (Euler’s p q phi function). Lemma 1. If xn 1 and xm 1,thenx m,n 1. “ “ p q “ Lemma 2. If x n,then xa n a, n . | |“ | |“ {p q Subgroups of cyclic groups Theorem Let H x . “x y 1. Every subgroup of a cyclic group is itself cyclic. (Specifically, if K H,thenK 1 or K xd where d is § “t u “x y the smallest non-negative integer such that xd K.) P a b 2. If H ,thenforanydistincta, b Z 0, x x . | |“8 P • x y‰x y 3. If H n,thenforeverya n,thereisaunique subgroup of H|of|“ order a. | ( ) This subgroup is generated by xd where d n a. ˚ m m,n “ { ( ) For every m, x xp q . ˚ x y“x y Example: Consider Z12 x .ThistheoremsaysthatZ12 has “x y exactly one subgroup of order a 1, 2, 3, 4, 6,and12.Theyare “ a 1: 1,a2: x6 ,a3: x4 x8 , “ “ x y “ x y“x y a 4: x3 x9 ,a6: x2 x10 , “ x y“x y “ x y“x y a 12 : x x5 x7 x11 . “ x y“x y“x y“x y Subgroup lattices A subgroup lattice is a graph whose nodes are the subgroups of a group G, with an edge drawn between every pair of subgroups H, K satisfying H K and there are no H L K. § ¨ ¨ Generally, we draw bigger subgroups higher than lower subgroups. Example: For G Z12 x , we saw the subgroups are given by “ “x y 1, x6 , x4 , x3 , x2 , and x G. x y x y x y x y x y“ So the subgroup lattice for Z12 will have 6 vertices. (In general, Zn will have one subgroup/vertex for every divisor of n.) To draw the edges, you must compute the partial order of containments. (When justifying your work, show these computations!) A partial (subgroup) lattice is a graph of the relative containment relationships of a subset of the subgroups of G. , Comparing lattices lattice of Z9 x : lattice of Z3 Z3 x y : “x y ˆ “x yˆx y x x y x y x yˆx y x3 x 1 x, y 1 y x y x yˆ xp qy ˆx y 1 1 Conclusion: Z9 and Z3 Z3 are not isomorphic! ˆ Thm: If G H,thenG and H will have the same lattice. – The reverse is not true (consider the lattice of Z4). Generators and relations: two ways Let G be a group, and let A be a subset of G. The relations are inherited from G,butwhatdoesA generate? We’ve seen: A generates the set of elements in G ... ¯ ✏1 ✏2 ✏` Let A a1 a2 a ai A, ✏i 1, ` Z 0 . “t ¨¨¨ ` | P “˘ P • u New definition: Consider all subgroups H G that contain A. § Define A is the smallest of these subgroups,i.e. x y A A , and if A H then A H. Ñ x y Ñ x y § (Does such a thing exist??) ¯ ✏1 ✏2 ✏` For reference: A a1 a2 a ai A, ✏i 1, ` Z 0 . “t ¨¨¨ ` | P “˘ P • u Proposition If is a non-empty collection of subgroups of G,thenthe S intersection of all members of is also a subgroup of G,i.e. S H G. § H £PS Definition If A is a non-empty subset of G, define the subgroup of G generated by A as A H. x y“ A H HÑ G £§ Proposition 1 In G, the set of elements generated as words in ai,a´ ai A t i | P u is the same as the smallest subgroup of G containing A,i.e. A¯ A . “x y Defining homomorphisms via generators Suppose G A and H are groups. When does a function “x y ' : A H extend to a homomorphism ' : G H? Namely, you Ñ Ñ know ¯ ✏1 ✏2 ✏` G A a1 a2 a ai A, ✏i 1, ` Z 0 . “ “t ¨¨¨ ` | P “˘ P • u So if you know 1. the images of ai for all ai A;and P 2. ' is a homomorphism; you know how to calculate ✏1 ✏2 ✏` ✏1 ✏2 ✏` ' a a a ' a1 ' a2 ' a . p 1 2 ¨¨¨ ` q“ p q p q ¨¨¨ p `q But is this map well-defined? 1. ' G H? Yes! H is a group. p q Ñ 2. Independent of representatives? Maybe! You need to check that the relations hold! Defining homomorphisms via generators Define ' on A and extend via ✏1 ✏2 ✏` ✏1 ✏2 ✏` ' a a a ' a1 ' a2 ' a . p 1 2 ¨¨¨ ` q“ p q p q ¨¨¨ p `q Independent of representatives? Check relations... Example: We know 2 2 S3 12 , 23 12 1, 23 1, 12 23 12 23 12 23 . “xp q p q|p q “ p q “ p qp qp q“p qp qp qy Which of the following extend to homomorphisms? 1. ' : S3 D6 defined by 12 s, 23 sr Ñ p q fiÑ p q fiÑ 2. ' : S3 S5 defined by 12 12 , 23 23 Ñ p q fiÑp q p q fiÑp q 3. ' : S3 S2 defined by 12 12 , 23 1 Ñ p q fiÑp q p q fiÑ 4. ' : S3 Z2 x defined by 12 x, 23 x Ñ “x y p q fiÑ p q fiÑ.

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