Second Course in Abstract Algebra
Second Course in Abstract Algebra MATH 114
Popular in Course
Popular in Mathematics (M)
This 3 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 114 at University of California - Berkeley taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/226606/math-114-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.
Reviews for Second Course in Abstract Algebra
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 10/22/15
SYLOW THEOREMS MATH 114 Let G be a nite group and lGl pnq7 where p is prime and does not divide q A subgroup of order p is called a Sylow p subgroup Theorem 01 There exists a Sylow p subgroup Proof Proof goes by induction on For lGl p the statement is trivial We assume that the statement is true for any group of order strictly less than First7 assume that G is abelian Let g E G and g 31 1 Let the order of g be pmb where p does not divide b First7 assume that m gt 0 Let N be the subgroup generated by 91 Then lNl pm and lGNl pn mq By induction assumption there exists a subgroup H in GN of order pn m Consider the natural projection 7T G a GN The preimage W l P is a subgroup in G of order p Now assume that m 0 Let U be the subgroup generated by 9 Then p does not divide By induction assumption GU has a Sylow subgroup of order p Hence there exists u E GU of order pk for some k gt 0 Again consider the natural projection 7T G a GU Let h E G be such that 7Th u Then the order of h is pkd Now we can repeat the argument in the previous paragraph with g h The case of abelian G is done Now let G be not abelian Let 017 70k be all the conjugacy classes in G Assume that 01 We have M lel lGl p q There is 239 gt 1 such that is not divisible by p First7 assume that gt 1 Pick up z 6 cl and let Gwg Glgzg 1z Since lCillel lGl an and p does not divide we obtain lel 10W for some b lt q By induction assumption G1 has a subgroup P of order p Then P is a Sylow p subgroup of G Now assume that we can not nd 01 such that 31 1 and p does not divide All conjugacy classes of order 1 form the center ZG of G If 05705117 70k are conjugacy classes of order gt 17 then M p q losl W WG l Date February 17 2006 2 SYLOW THEOREMS MATH 114 By our assumption p divides for all 239 5k Therefore p divides lZG Thus7 we obtain that lZGl pkc for some k gt 07 p does not divide c By induction assumption one can nd a subgroup N in Z G of order pk Note that N is a normal subgroup of G Again by induction assumption there is a subgroup H in GN of order pn k Consider the natural projection 7T G a GN The subgroup P W l has order p Theorem is proven D Theorem 02 The number of Sylow p subgroup is congruent to 1 modulo p Proof Let 9 denote the set of all subgroups of G of order p Then G acts on Q by conjugation Let P E 9 de ne the subgroup NP9 Glng 1P Then the order ofthe G orbit of P equals Mfg By de nition P is a normal subgroup of N Lemma 03 Let P7 P E 9 Suppose that P C NP Then P P Proof P is normal in NP By the second isomorphism theorem P P P PP P Let lPP P l p 7 then lP P l 10 Therefore a 07 P P P 7 that immediately implies P P D Consider now P action on Q by conjugation Then every P orbit has p5 elements and exactly one orbit has 1 element lndeed7 let G be a P orbit of some P E 9 Then W o l lNP oPl Since lPl 10 0 105 for some 5 lf 0 17 then P C NP and by Lemma 03 P P Since the number of elements in Q is the sum of orders of all P orbits7 we obtain 9 E 1 mod p Theorem is proven D Theorem 04 All Sylow p subgroups are conjugate In other words ifP and P are two Sylow p subgroups then P ng l for some 9 E G In particular all Sylow p subgroups are isomorphic Proof We have to show that Q has exactly one G orbit Assume that 9 is a G orbit of some P E Q It was proven above that all P orbits have order p5 and there is only one P orbit P of order 1 But the order of 9 is the sum of orders of P orbits in 9 Hence lQl E 1 mod p SYLOW THEOREMS MATH 114 3 Suppose that Q 31 9 Then there is Q E 9 such that Q 9 But then every Q orbit in 9 has order p5 with s 2 1 Then we have 9quot E 0 mod p We obtain a contradiction Hence our assumption 9 31 9 is wrong D Corollary 05 The number of SyIOW p subgroups divides
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'