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# AbstractAlgebraI MATH533

Drexel

GPA 3.63

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This 15 page Class Notes was uploaded by Moshe Swift III on Wednesday September 23, 2015. The Class Notes belongs to MATH533 at Drexel University taught by Staff in Fall. Since its upload, it has received 25 views. For similar materials see /class/212304/math533-drexel-university in Mathematics (M) at Drexel University.

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Date Created: 09/23/15

NOTES ON THE PROOF OF THE SYLOW THEOREMS 1 The Theorems We recall a result we saw two weeks ago Theorem 11 Cauchy s Theorem for Abelian Groups Let A be a nite abelian group Up is a prime number that divides its order then A must have an element of order p Theorem 12 Sylow s First Theorem Let G be a nite group Let p be a prime number such that its power by oz is the largest power that will divide Then there exists at least one subgroup of order po Such subgroups are called Sylow psubgroups PROOF We divide the proof into two cases Case One p divides the order of the center ZG of C By Cauchy s Theorem for abelian groups ZC must have an element of order p say a By induction the quotient group C lt a gt must have a subgroup P0 of order pa l Then the preimage of P0 in ZC is the desired subgroup of order po Note in general if S is any subset of a quotient group CH then the order of the preimage of S is the product of its order with the order of the subgroup Case Two assume that p does not divide the order of the center of G Write 0 in terms of the class equation77 lGl lZGl ZlCOH alla where the sum is over all the distinct noncentral conjugacy classes of C that is conjugacy class with more than one element Since p fails to divide the order of the center there must be at least one noncentral conjugacy class say Conjb whose order is not divisible by p Recall that lConjbl G Cab lCllCGbl We observe immediately that po must divide the order of the subgroup Cab Again by induction G will have a Sylow psubgroup This ends the proof Corollary 11 There is a subgroup Q of a Sylow psubgroup P for every power ofp that divides the order of the group C Corollary 12 If every element of a group is a power of a prime p then the group is a pgroup that is the order of the group is a power ofp Theorem 13 Sylow s Second Theorem Let np be the number of Sylow psubgroups of a nite group G Then np E 1 mod p PROOF We begin with a claim Claim Let P be any Sylow psubgroup lfg E G be a p element and ng 1 P then 9 E P To see this consider the subgroup R generated by g and P By assumption 9 E NaP so R S NaP Hence P is a normal subgroup of R We nd lRl lRPl But lRPl is a cyclic group generated by the coset gP Then gP is a pelement since 9 is Hence lRl is a power ofp since all its elements are pelements Let Sp be the set of all Sylow psubgroups of G Then G acts on this set by conjugation Let P Q E Sp be two distinct subgroups Then Q cannot be xed under conjugation by all the elements of P because of the Claim Let 9 be the Porbit of Q under conjugation Then the size of the orbit must be divisible by p because of the orderstabilizer equation P StabPQ Since P is a power of p the size of any orbit must be a power of p The case 0 p0 1 is ruled out since Q cannot be xed by all the elements of P We nd that the set of all Sylow psubgroups is the union of Porbits There is only one orbit of order one P while the other orbits must have orders a positive power of p We conclude np Sp E 1 mod p 0 Remark We want to emphasize a result from this proof Let P be any Sylow p subgroup As above we let P act on Sp by conjugation Let So be any Pinvariant subset of Sp which means that is a disjoint union of Porbits Then So E 0 modp if P SO while So E 1 modp if P E SO Theorem 14 Any two Sylou psubgroups dre conjugate PROOF Let P be any Sylow psubgroup Let So be the set of all Gconjugates of P Then So is Pinvariant and P E SO By the above observation SO E 1 mod p If SO does not exhaust the set of all Sylow psubgroups choose one say Q not in So Let Sl be the set of all Gconjugates of By the same reasoning as for So with Q playing the role of P we must have Sl E 1 mod p On the other hand 1 is Pinvariant and P Sl By the above observation Sl E 0 mod p Contradiction Corollary 13 The number np of Sylou psubgroups must divide G po PROOF Recall that the number of conjugates of any subgroup H in a group G is given by conjugates i NGH If H is a Sylow psubgroup then the order of its normalizer must be divisible by po since H 3 Na But the number of all Sylow psubgroups is just the number of conjugates of any one of them Theorem 15 Any psubgroup B is contained in d Sylou psubgroup PROOF Let B act on the space Sp by conjugation Then the size of any Borbit 9 must be a power of p since the O G NaB Since the size of 8 is not a power of p there must be at least one Borbit with one element say P But B must be a subgroup of P since the subgroup generated by B and P is a power of p by the corollary to Sylow s First Theorem Theorem 16 Any nite dbelidn group is a product of its Sylou psubgroups As a consequence of this last theorem to classify the nite abelian groups it is enough to understand their structure in the case that their order of a power of a prime One can show that if A p then A is isomorphic to a group of the form Zn1 gtlt Zn2 gtlt Zn where n1 2 n2 2 2 me and n1 nk n 2 Other Theorems 1 close with the statement of three theorems of Burnside Their proofs is beyond the scope of the course Theorem 21 Let P be a Sylow psubgroup of a nite group G IfP is central in its own normalizer then G has a normal subgroup U such that G P U and P O U e Theorem 22 Let G be a nite group of order N in which every Sylow subgroup is cyclic Then G is generated by two elements A and B with the relations AmB 1BAB ATNmn GCDr 71nm1rn E lmod in Theorem 23 If a nite group G has a conjugacy class whose size is pk where p is a prime and k 2 1 then G cannot be a simple group 3 SemiDirect Products Suppose G A Q where A is normal in Gt Then we can work with G as ordered pairs 111111 39 112112 a141a242 11111a211f1q1112 So we can de ne the multiplication on ordered pairs as 11111039 112112 a1a41a241q2 where aqx qxq ll The inverse of a q is given by aq 1a 1 1 1 Note that we can treat oz as a homorphism from the group Q into the automorphism of Al In fact this is all we need to de ne such a group G we are given two groups A and Q and a homomorphism from Q into AutAl The resulting group is called the semidirect product of Q with Al As a consequence it is now important to understand the automorphism groups of some low order abelian groups because it is through them that many nonabelian groups may be constructed It is useful to verify that a 1Q a E A is a normal subgroup of the semidirect product isomorphic to A while IA 1 q E is a subgroup of the semidirect product isomorphic to Furthermore we can verify that lAaqa1Q1Aaq71 aqa1Q We give a brief review of some automorphism groups of some abelian groups For Zp where p is a prime then AutZp E Z that is the group of positive integers less than p under multiplication modulo pl For Zm where m is composite then AutZm E UZm that is the group of positive integers less than in and relatively prime to in under multiplication modulo ml For example AutZ4 is 1 3 and is isomorphic to Z2 while AutZ6 is 1 5 and is also isomorphic to Z2 For C Zp gtlt Zp where p is prime we nd that AutG GL2 Z17 Note one way to give a homomorphism of a cyclic group Zn into any of the above automorphism groups is to nd an explicit automorphism of order n in the automorphism group itself Further in forming such as homomorphism we also note the basic fact that if lt1 G1 A G2 is any homomorphism between two groups and x E 01 then the order E G2 must divide lG2l as well as Hence the order of must be a common divisor of lGll and lG2l Examples 1 P90quot 01 T1 90 The dihedral groups D7 are all semidirect products where A Zn and Q Z2 0 l with the map aqx Elqx where x 0 or 1 We may also write ozlx n E l 95 Let x be a generator of A Zn and write a x 0 and b 0 1 We nd ab x o1gtltxalt1gtoolgtltz1gt ba 0 1 9600a1I10I0 In particular a and b satisfy the relations for the dihedral group a e b2 e and ab ba l Since AutZn is isomorphic to Un we may readily generalize this example Let A Z2 gtlt Z2 and Q S3 Note AutZ2 gtlt Z2 E S3 The resulting group has order of 24 Let F be a eld then we can form the semidirect product of F with Fquot with af x f x wheref E Fquot and 95E F f13x139f2x2f1 01951f2 961962 Consider the semidirect product of F with GLn F by flaT139 f2 T2 f1 T1f1T1T2 Note with n 2 and F Z2 we obtain another group of order 24 We can construct a group G of order 20 by a semidirect product of Z5 with Z4 because AutZ5 E Z4 11151 39 11252 a1 0450112 b1 52 Note C has generators x and y satisfying x5 y4 e and xyx y This example is a special case of the procedure in 4 with F Z5 Another realization of this group is the subgroup of S5 generated by the two cycles p l 2 34 5 with a 2 3 54 so that the conjugate of p by a is a power of p One can show that the group of order eight of unit quaternions cannot be expressed as a semidirect product Examples 4 and 6 are particular instances of a metacyclic group which is given as the semidirect product of Zn with its automorphism group Note that the metacyclic group can be viewed as a subgroup of the symmetric group Sn We will see this group later in our study of Galois theory in connection with the roots of x E 0 9 There is a canonical semidirect product associated to any group G Let n gt 0 Let Sn act on the product group 7 G X X X G ntimes by permuting coordinates that is if a E Sn let 7 x1x2 xn 9551 9502 The resulting semidirect product group is called the Wreath product of Sn with C It has order n The group of signed permutation matrices and the symmetry group of the cube are examples of wreath products H O The group G of signed n X n matrices can be expressed as a semidirect product of the normal subgroup of all diagonal matrices diagi1 i1 i1 where the signs are chosen indepen dently and the subgroup of all permutation matrices The normal subgroup of G consisting of all matrices of determinant 1 gives the group of rotations of the cube in nspace It is a group of order 2n 1nl This is another natural family of nonabelian groups 4 Sylow Theory and Classi cation of Low Order Groups We begin by stating some basic results about the normality of Sylow psubgroups These statements all follow easily from the techniques described introducted the last lecture Below we shall re ne these techniques to give an actual classi cation of all groups of a given order 1 If G pq where p lt q are distinct primes then a ifp q 7 1 then there are unique Sylow p and 1 subgroups b ifp q 7 1 then there is a unique Sylow q subgroup only F If G pqr where p lt q lt r are distinct primes then a the Sylow rsubgroup is normal b C has a normal subgroup of order qr c if q r 7 1 then the Sylow qsubgroup of G is normal 9quot If G p211 where p and q are primes then either the Sylow p or Sylow 1 subgroups are norma It is useful to experiment with integers that have the above the prime factorizations 5 Classi cation of Groups 12 We will move onto the classi cation of groups of order 12 The groups we know are two abelian Z2 gtlt Z2 gtlt Z3 Z4 gtlt Z3 and two nonabelian the alternating group A4 and the dihedral group DG E S3 gtlt Z2 Further last week we characterized A4 among the groups of order 12 as having 713 4 In fact we saw that if G is any group of order 12 with 713 4 then there was a homomorphism of G into the group of permutations of the set of the four Sylow 3subgroups By considering the elements of order 3 and counting we found that the image of G in S4 was contained in A4 Since they have the same order they must be equal It is now easy to classify the abelian groups A of order 12 By Sylow theory we know that A has two unique subgroups P3 of order 3 and P2 of order 4 By the recognition of direct products theorem we nd A E P3 gtlt P2 Recall that we have already classi ed the abelian subgroups of order 3 and 4 They are for order 3 Z3 while for order 4 Z4 or Z2 gtlt Z2 Note Z12 E Z4 gtlt Z3 It remains to examine the case 713 1 for a nonabelian group of order 12 Let P3 be the unique Sylow 3subgroup which must be normal Let P2 be some Sylow 2subgroup Then G P3 P2 that is G must be a semidirect product of P2 with the cyclic group of order 3 Z3 Further AutZ3 E Z that is it is isomorphic to 1 2 under multiplication modulo 3 So there is a only one nontrivial automorphism say oz of order 2 given by x gt 295 for x E Z3 There are two main cases 1 For the rst case P2 Z2 gtlt Z2 We need to consider homomorphisms 9 from P2 into the automorphism group of Z3 which itself is isomorphic to Z2 There are three nontrivial homomorphisms 91 239 1 2 3 To construct them let a b c be the nonidentity elements of P2 We nd 9111 oz while 911 91c ld 921 oz while 921 92c Id 93 c oz while 9311 931 Id The resulting semidirect product group is isomorphic to DG E 53 X Z2 2 For the second remaining case P2 E Z4 Then for a nontrivial homomorphism 9 must map the generator 1 of Z4 onto the a The resulting nonabelian group is not isomorphic to the ones consider already To conclude there are three nonabelian groups of order 12 and two abelian ones 6 Classi cation of Groups of Order 30 Let G be a group of order 30 2 3 5 the product of three distinct primes Let np denote the number of distinct Sylow psubgroups of G Then we know that 71 E 1 modp and np must diVide lGlpe where p 3 is the maximal power ofp that diVides In our case we nd that 713 1 3k and must diVide 10 so 713 1 or 10 while 715 1 mod 5 and 715 must diVide 6 so 715 10r6 Now either 713 or 715 must equal 1 for otherwise G would contain 20 elements of order 3 and 24 elements of order 5 Let P3 be a Sylow 3subgroup and P5 be a Sylow 5subgroup We know that one of them must be a normal subgroup As a consequence the product set H P3 P5 is in fact a subgroup of G of order Because the index G H 2 we nd that H is a normal subgroup of G Further by our comments last time about the structure of groups whose orders are the product of two distinct primes we know that H E Z3 gtlt Z5 since 3 K5 71 4 Since H is normal in G G H P2 where P2 is some Sylow 2subgroup of order 2 To sum up to determine the structure of groups of order 30 we only determine the number of possible semidirect products of Z3 gtlt Z5 with Z2 Let 9 Z2 A AutZ3 gtlt Z5 We may easily verify that AutZ3 gtlt Z5 E AutZ3 gtlt AutZ5 since Z3 and Z5 are simple groups of distinct orders In particular 9 becomes a homomorphism from Z2 to UZ3 gtlt UZ5 or Z2 A Z2 gtlt Z4 Let a be a generator of UZ3 E Z2 b a generator for UZ5 E Z4 while c a generator for Z2 Then there are exactly four choices for 9 they are aaa aaa aaail aaail bab babil bab babil We nd that the resulting semidirect product groups are all nonisomorphic by considering their centers The order of the centers from left to right are 30 3 generated by a 5 generated by b and 1 Of course if the center has order 30 then the group is abelian 7 Classi cation of Groups of Order 28 Let G be a group of order 28 4 7 22 7 Let np denote the number of Sylow p subgroups as usual Then 717 E 1 mod 7 and 717 must divide 4 Hence n7 1 and the only Sylow 7subgroup P7 is normal and isomorphic to Z7 Let P2 be a Sylow 2subgroup of order 4 Then P2 may be isomorphic to either group of order 4 Z4 or Z2 gtlt Z2 Since P7 is normal in G the group G P P2 that is a semidirect product of P7 with P2 Hence we need to classify these semidirect products or equivalently the homomorphisms 9 from P2 into AutP7 in particular 9 Z4 A UZ7 9 Z2 gtlt Z2 A UZ7 where AutZ7 E UZ7 the multiplicative group of integers relatively prime to 7 Note U7 E Z6 1 We rst study the case that the Sylow 2subgroup is cyclic to isomorphic to Z4 Write Z4 lt a gt and b 9a E 17 Then lbl is a common divisor of 4 and 6 The only choices for its order are 1 and 2 1f lbl 1 G is simply the direct product of Z4 and Z7 When lbl 2 we need to nd the elements of order 2 in U7 1 2 3456 under mul tiplication modulo 7 There is only one such element 6 Hence there is exactly one such homomorphism 9 Z4 A U7 given by 995 695 E 17 F If the Sylow 2subgroup is isomorphic to Z2 gtlt Z2 then the needed homomorphism has the form enzngQEUm As in the last case the trivial homomorphism yields the direct product group Z2 gtlt Z2 gtlt Z7 For 9 to be nontrivial we determine its actions on the generators a1 1012 0 1 and a3 1 1 in Z2 gtlt Z2 Now exactly two of the three generators may map to 6 while the third maps to 1 However each such choice for 9 gives an isomorphic group We conclude that there are four groups of order 28 of which 2 are abelian and 2 are nonabelian 8 Comments on Groups of Order 60 Let G be a group of order 60 22 3 5 Then G has Sylow psubgroups of order 4 3 and 5 We can show that there are 13 distinct isomorphism types of groups of order 60 81 G is simple Assume G is a simple group of order 60 Then n5 6 and n3 10 By simplicity 713 gt 1 and n5 gt 1 Now 715 E 1 mod 5 and 715112 which implies 715 must equal 6 For 713 we similarly nd 713 E 1 mod 3 and 713120 Hence n3 equals either 4 or 10 Suppose n3 4 so there are 4 Syow 3subgroups forming the set 83 Then G acts transitively on 83 which gives a nontrivial homomorphism oz from G into S4 Since G is simple oz must be 11 This is impossible since 24 1514 lt 0 60 Hence 713 must equal 10 CLAIM 1 if G has a subgroup of order 12 then G A5 To verify the claim consider the Gaction on the coset space GH This action gives a nontrivial homomorphism 6 from G into S5 Since G is simple 6 must be 11 Since A5 is a normal subgroup in S5 we nd A5 0 must be a normal subgroup in A5 Since A5 is simple must equal A5 itself Hence the Claim is established CLAIM 11 G has a subgroup of order 12 By simplicity n gt 1 By Sylow theory 712 E 1 mod 2 and n2l15 which implies 712 must equal 5 or 15 1f 71 5 then we may let G act on the set 82 of Sylow 2subgroups This gives a nontrivial homomorphism y from G into S5 By simplicity this map is 11 Arguing as in Claim 1 we nd G E A5 If n 15 there must be two Sylow 2subgroups say P and Q with an intersection of order 2 To see this suppose all 15 Sylow 2subgroups intersect only at the identity This gives 45 elements of even order However G has 24 elements of order 5 and 20 elements of order 3 since n5 6 an 713 10 This is impossible so there are such subgroups P and Let x be a nonidentity element from P O Of course 95 cannot be a generator for either P or Q so it has order 2 Consider the centralizer Cgx Clearly both P and Q must lie in the centralizer of 95 So 4 must be a divisor of the order of CG Furthermore by examining the elements of P and Q we nd Cgx must contain at least six distinct elements as well Hence ngx is a common multiple of 4 and 6 and be a divisor of 60 as well There is only one such integer namely 12 Note one can show that if 715 gt 1 then G is automatically simple 82 Sylow 3 and 5 subgroups are normal We shall assume that the Sylow 3subgroups P3 and 5subgroups P5 are unique and normal Then N P3 P5 is a subgroup of G Further N is a normal subgroup since gNg 1 ngg 1 gP5g 1 P3 P5 for any 9 E G Of course N E Z15 and AutZ15 UZ15 which is isomorphic to the set 1 24 7 8 11 13 14 under multiplication modulo 15 So UZ15 Z4 gtlt Z2 To sum up G NH where H is a subgroup of order 4 We nd eleven isomorphism types 1 H E Z4 We list ve distinct homomorphisms 9 from Z4 into Z4 gtlt Z2 a 91 0 0 b 01 10 or 01 3 0 C 91 20 61 91 01 8 91 11 21 H E Z2 gtlt Z2 We list six distinct homomorphisms 9 from Z2 gtlt Z2 into Z4 gtlt Z21 a 01 0 0 0 and 00 1 00 b 01 0 20 and 00 1 0 0 c 010 01 and 00 1 0 0 d 01 0 20 and 1901 21 e 010 01 and 1901 21 f 010 21 and 00 1 0 0 9 Comments on the Simple Group of Order 168 Recall that the general linear group GL2 01 over a eld With 11 elements has order q2 71q2 iq lts normal subgroup the special linear group SL2 q has order qq1q71 since det GL2 q A FX With kernel precisely SL2 q Finally the center of SL2 q is i1 The resulting quotient group the projective linear group PSL2 q has order qq 1 1 12 We are going to work With the case 11 71 Then SL2 7 has order 336 and PSL2 7 has order 1681 A convenient feature of the eld of order 7 Z7 is that its quadratic extension can be realized as the Z7239 that is the ring of Gaussian integers m in mn E Z With usual complex addition and multiplication considered modulo 7 We Will nd the order of each conjugacy class in SL2 7 to begin We present the table of results then give comments Conjugacy Classes for SL2 7 J2 14 24 7 J2 14 24 In order to compute the size of a conjugacy class of an element x it is enough to nd the order of the centralizer of x g E G gar mg The table entries are arranged to minimize the number of such calculations It is plain to observe that i1 iD have the same centralizers Further J and i51 have the same centralizers as well as the four elements iJl and iJQ To nd the centralizer of a a c results A further restriction needs to be applied detg 1 It is interesting to note that while J1 and 7 ave the same spectral data 1 is the only eigenvalue with multiplicity 1 so are conjugate in the full linear group they are not conjugate in the special linear group We next consider how the conjugacy classes in SL2 7 relate to those of PSL2 7 SL2 7il We begin with noticing that g E C601 ltgt 79 E Caa where C601 is the centralizer of a in the group G SL2 7 In particular the size of the set C601 is halved when sent to its image in the quotient group On the other hand a and 711 have the same image in the quotient Hence the images of the following pairs agree i1 iDiJ1iJ2 and i5I J Hence the 11 conjugacy class of SL2 7 collapse to 6 If 7r SL2 7 A PSL2 7 be the canonical projection then le5L7ral lCSLal2 lCSLEal2 where a E iIiDiJ1iJ2 i5I J that is a y J With these remarks we have the following table for PSL2 7 We use the notation a to denote the class of an element of a in the quotient matrix as we consider 9 d and write gar avg A linear system for the entries 1 b cd Conjugacy Classes for PSL2 7 Matrix Order lConjugacyClassl I 1 1 D 3 56 J 2 21 J1 7 24 51 J 4 42 J2 7 24 The order of the elements in the above table follow from observing D3 71 J17 J 71 and 5H J4 711 We are now able to show that PSL2 7 is simple by the same elementary counting argument that we had used for the alternating group A51 Let N be a normal subgroup of PSL271 Then if a E N so must its entire conjugacy class That forces the order of N to have the form lNl 156a21b24c42d245 where a bc 515 may equal either 0 or 1 Recall the basic fact that lNl must be a divisor of 1681 By casebycase analysis we nd that there are no solutions for a b c 515 that satisfy these two conditions other than lNl 1 1681 To verify this it is convenient to separate out the two cases that lNl is odd or even We conclude that PSL2 7 is simple We close by counting the Sylow psubgroups of PSL271 Note 168 23 3 71 Further we recall that np PSL2 7 NaPp where P17 is the Sylow psubgroup Sylow psub39r0ups P for G PSL2 7 Prime p Number 71 lNG 2 21 8 3 28 6 7 8 21 The number of Sylow subgroups for primes 3 and 7 follow immediately from counting the elements of those orders in the group itself The count for the Sylow 2subgroups though requires a special calculation that requires some wor 1 If np is the number of Sylow psubgroups then we nd np E 1modp npl168 as predicted by the Sylow theory One can show that P2 is isomorphic to the dihedral group D4 of order 8 NaP3 is isomorphic to S3 and NaP7 has order 211 To verify the results for p 3 7 observe that PSL2 7 has no elements of order 6 nor 211 Hence the corresponding normalizers cannot be abelian by the classi cation of groups of order 6 and 211 10 Detailed Information for Some Low Order Groups 101 Symmetric Group S4 Conjugacy Class Elements Order 4 6 4 31 8 3 22 3 2 212 6 2 14 1 1 Wow 2 3 D4 34 S3 Notes The SyloW 2subgroups are generated by the conjugacy class 22 and by a certain transpo sition and a certain 4cycle 1 12 or 34 it contains 1324 2 13 or 24 it contains 1234 3 14 or 23 it contains 1342 The SyloW 3subgroup is generated by some element of order 3 say 123 Its normalizer contains the transpositions 12 23 and 13 102 Alternating Group A4 Conjugacy Class Elements Order 31 8 3 22 3 2 14 1 1 20 Wow 4 2 1 A 3 4 S3 Comment A4 has no subgroup of order 6 103 Symmetric Group S5 Conjugacy Class Elements Order 5 24 5 41 30 4 312 20 3 32 20 6 221 15 2 213 10 2 15 1 1 p Sp NGP Comment 2 15 8 D4 3 10 6 S3 5 6 10 D5 Comments Sylow 2subgroup P2 4524x3525x3423gt45gtlt2435gt2453103 0 it equals its oWn normalizer SyloW 3subgroup P3 lt 235 253 gt normalizer has order 6 and equals 1423a1435a 235 25311425a235a 253 0 SyloW 5subgroup P5 lt 12345 gt normalizer has order 10 and equals 1345a 14x23 25x34 15x24 12x35 12345 1 104 Alternating Group A5 Conjugacy Class Elements Order 5Lr 12 5 5 12 5 312 20 3 221 15 2 15 1 1 p Sp NGP Comment 2 5 12 A4 3 10 6 S3 5 6 10 D5 Comment SyloW 2subgroup P2 is generated by elements from the conjugacy class 221 for example P2 1245 1425 1534 lts normalizer has order 12 and is given by 14313451531451245a1425a1534a The normalizer is generated by a copies of Z2 gtlt Z2 and Z3 13 11 More Comments on Group Classi cation The main feature of the groups that we can successfully analyze have orders whose prime factoriza tions involve at most three distinct primes with low powers and can be expressed as a semidirect product It is also within the scope of our techniques to classify the groups of order p3 where p is a prime In contrast there are 2328 distinct groups of order 27 128 12 Last Remarks Yet some more comments about the state of group theory in the late 19th and early 20th centuries from an article by T Y Lam of Berkeley that appeared in the Notices of the American Mathematical Society 1998 By the 1890s the known simple groups were the alternating groups A7 with n 2 5 Jordan s projective special linear groups PSL2p with n 2 5 and some socalled sporadic simple groups The American mathematician FiN Cole found a simple group of order 504 which was recognized 14 later as PSL28 that is over a nite eld With 23 8 elements The German mathematician Holder found all simple groups of order less than 200 in 1892 Cole to order 660 in 1893 and the British mathematician Burnside to order 1092 In this era Burnside also showed that if the order of a simple groups is even then it must be divisible by 12 16 or 56

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