Math 31 Lesson Plan. Day 27: Fundamental Theorem of Finite Abelian Groups. Elizabeth Gillaspy. November 11, 2011


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1 Math 31 Lesson Plan Day 27: Fundamental Theorem of Fnte Abelan Groups Elzabeth Gllaspy November 11, 2011 Supples needed: Colored chal Quzzes Homewor 4 envelopes: evals, HW, presentaton rubrcs, * probs Goals for Students: Students wll: Understand the statement and applcatons of the Fund Thm of Fnte Abelan Groups 1
2 [Lecture Notes: Wrte everythng n blue, and every equaton, on the board. [Square bracets] ndcate antcpated student responses. Italcs are nstructons to myself.] Return quzzes, HW 5. collect homewor 6 I ve been hntng at ths Fundamental Theorem of Fnte Abelan Groups for most of the term, and now we fnally meet t. We re not gong to dscuss the proof n class you can read t n the textboo, or n Gallan, f you re so nclned but snce t s such a cool (and useful!) theorem I wanted to at least state t. I thought t would be a nce way to end the group theory part of ths course. Whch means that we ll be startng wth rng theory on Monday or perhaps today, f we have tme. Are there any questons about Secton 13, or the homewor, before we get started? I thn the Fundamental Theorem wll mae most sense f we start wth an Example: How many dfferent abelan groups are there of sze 180? To fgure ths out, we start by factorng 180: 180 = The Fundamental Theorem tells us that the possble groups of order 180 are Z 2 Z 2 Z 3 Z 3 Z 5 Z 4 Z 3 Z 3 Z 5 Z 2 Z 2 Z 9 Z 5 Z 4 Z 9 Z 5 Can anyone tell me how I fgured ths out? don t wat too long for an answer Snce 180 = , f we want to wrte a group of sze 180 as a drect product, we wll 2
3 need a factor of sze 4, a factor of sze 9, and a factor of sze 5. There are two possble 2groups (groups whose order s a power of 2) of sze 2 2 = 4 Z 2 Z 2 and Z 4. Smlarly, there are two possble 3groups of sze 3 2 = 9.] How many 5groups are there of sze 5? [one] Snce there s only one possble 5group of sze 5, the groups lsted above are the only ways to wrte 180 as a drect product of cyclc groups. The Fundamental Theorem tells us that every abelan group s somorphc to a drect product of cyclc groups. In other words, we don t have to thn of any other possbltes for abelan groups of sze 180 nothng more complcated than drect products! Ths s most defntely NOT true n the nonabelan case. People have managed to classfy all fnte smple groups abelan and not but ths was a massve undertang! To get a better understandng of the Fundamental Theorem, we need some more tools e, defntons. Def: If p s a prme, a ppower group s a group G such that G = p for some postve nteger. How many abelan groups of sze p are there n general? Def If s a postve nteger, we say that the number of parttons of, wrtten p(), s the number of ways to wrte as a sum of postve ntegers. (Order doesn t matter.) For example, p(4) = 5. Grab a partner and try to fnd all 5 parttons! Parttons of 4: ; ; 1 + 3; 2 + 2; 4. The number of abelan groups of sze p s the number of parttons of. So, how many abelan groups are there of sze 3 4 = 81? What are they? Wor wth your partner to fgure t out. Once you now one, please come wrte t on the board. 3
4 The abelan groups of sze 3 4 are: Z 3 Z 3 Z 3 Z 3 ; Z 3 Z 3 Z 9 ; Z 3 Z 27 Z 9 Z 9 Z 81 We say that order doesn t matter for parttons, because order doesn t matter for drect products. A B = B A always. Algorthm for fndng all the abelan groups of order n: 1. Factor n nto prmes: n = p r 1 1 p r p rm m. 2. Fnd all the parttons of r Fnd all the abelan groups of sze p r 1 1 there wll be p(r 1 ) of them. 4. Repeat for all the prme factors of n. 5. Buld drect products of these p groups n all the ways possble to end up wth a group of sze n. Questons about ths algorthm? Please get nto groups of 3 or 4. Try to mae sure there s one person n your group that you haven t wored wth before. Use the Fundamental Theorem (e, the algorthm above) to fnd all the abelan groups of the followng szes:
5 = = once everyone has fnshed the frst two, dscuss at board. Prme factorzaton of 168 : Parttons of 3: , 1 + 2, 3. Abelan groups of sze 168: Z 2 Z 2 Z 2 Z 3 Z 7 Z 4 Z 2 Z 3 Z 7 Z 8 Z 3 Z 7 Prme factorzaton of 1125 : Parttons of 2: 1 + 1, 2. Abelan groups of sze 1125: Z 3 Z 3 Z 5 Z 5 Z 5 Z 3 Z 3 Z 25 Z 5 Z 3 Z 3 Z 125 Z 9 Z 5 Z 5 Z 5 Z 9 Z 25 Z 5 Z 9 Z 125 In none of these examples above (well, except 3 4 ) do we see Z n. However, we now that t s an abelan group of sze n. In the examples above, whch one s somorphc to Z n? Please try to fgure ths out n your groups. 5
6 [In the examples above, Z n s always somorphc to the last drect product n the lst. Example 2 on page 134 tells us that a cyclc group of order n s somorphc to the drect product of two cyclc subgroups of szes, l ff l = n and (, l) = 1. So, n general, the abelan group of sze n that s a drect product of cyclc pgroups s the one that s somorphc to Z n.] Are there more questons about the Fundamental Theorem of Fnte Abelan Groups? I want to tal brefly about Corollary Some of you mentoned that we ve seen theorems le t before. So let s put all these relevant statements on the board so we can compare: Corollary 14.5 states that f G s an abelan group of order n, and m n s any dvsor of n, then G has a subgroup of order m. Theorem 11.7 states that f G s an abelan group and p G s a prme, then G has a subgroup of sze p. Theorem 5.5 states that f G s a cyclc group of order n, then G has a unque subgroup of sze m for any m n. What are some advantages of Corollary 14.5 over our prevous results? Thnparshare Corollary 14.5 s cool because t doesn t requre ether that G be cyclc or that the dvsor of G be prme. Proof of Corollary 14.5: If G s an abelan group of sze n, then the Fundamental Theorem says t can be wrtten as a drect product of cyclc groups of prmepower order. In other words, we can wrte G = Z p r 1 1 Z p r Z p r, where the p s are all prmes but not necessarly all dfferent prmes. Note that snce G = n, we must have n = p r 1 1 p r p r. 6
7 If m n, then each of the prme factors of m s also a prme factor of n. Can the power of a prme factor of m be bgger than ts power n n? [no] Moreover, each prme factor of m s rased to a lower power n m than t s n n. In other words, we can wrte m = p s 1 1 p s p s, where 0 s r for all. Note that Z r p s cyclc for each, and also that p s Z p r, snce ps pr. Hence, by Theorem 5.5, we now that each Z r p has a cyclc subgroup of sze p s call t P. Then, P 1 P 2... P G = Z p r 1 1 Z p r Z p r, and snce P = p s we now that P 1 P 2... P = p s 1 1 p s p s = m. So we have found a subgroup of G of sze m, as desred. Do an example! Introduce rngs f tme 7
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