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by: Kavon Feest

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# Second Course in Abstract Algebra MATH 114

Marketplace > University of California - Berkeley > Mathematics (M) > MATH 114 > Second Course in Abstract Algebra
Kavon Feest

GPA 3.93

Staff

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COURSE
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KARMA
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## 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.

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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

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