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# ALGEBRA MATH370

Penn

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This 3 page Class Notes was uploaded by Claudine Friesen on Monday September 28, 2015. The Class Notes belongs to MATH370 at University of Pennsylvania taught by Staff in Fall. Since its upload, it has received 47 views. For similar materials see /class/215405/math370-university-of-pennsylvania in Mathematics (M) at University of Pennsylvania.

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Date Created: 09/28/15

Notes 928 Math 370 Recall Last time de ned left right cosets If H S G and G UaH a disjoint union then each left coset is in bijection with H when lGl lt 00 Then 11G H where the latter term is the number of left H cosets Example ZpZX for a xed prime p This is a nite group a group of units in a nite eld The group has p71 elements We also know that this is a eld since p is prime so every nonzero element is a unit We know that the order of each element in the group must divide p 7 1 from last class Consequently If you take anly element a E ZpZX then ap l E 1 mod p This is called Fermat7s little theorem Now how do we generalize this to ZnZX7 Answer Va 6 Z with an 1 aw 7 1 mod n Recall given a group G we de ned the notion of the conjugacy class of z in G This is the set ymy lly E G ie the set of all conjugates of z in G Last time showed there is a bijection ymy lly E G GZgz where ZGz is the centralizer of z in G ie ZGx g E Glgzg l And we showed that that the bijection associates aZGx ltgt na l To show that this is actually injective aza l bzb l ltgt b lasa lb z ltgt b la E ZGz ltgt bZGz aZGx Exercise show that this map is well de ned This is explicit in the reverse implication arrows ltgt De ne an equivalence relation two elements are equivalent if they are conjugates We know G is the disjoint union of all conjugacy classes in G that7s actually trivial When lGl lt 00 then this implies M G 21mm where z runs through a set of representatives of conjugacy classes in G Rephrased Let 1 5 be the set of all conjugacy classes in G Pick 6 513239 1771 Then there is a bijection Hence 1181 IiGIllZGWi N N G So 11G 15 Z 20 i1 i1 De nition let p be a prime number A p gmup is a nite group with cardinality pm for some m E N Lemma let a nontrivial group G ie G 31 be a p group Then ZG 31 e7 ie G has a non trivial center The proof follows immediately from the previous summation N G U Si the disjoint union of conjugacy classes i1 Why is this proof suf cient lf 1 pick an element x in the center of the group7 how many elements are in the conjugacy class containing x If an element is in the center7 then by def inition7 any conjugation of that element yields itself Hence by de nition7 centers of groups are composed of elements whose conjugacy classes contain just themselves Say 51 6 Want to show 1 for some 2 S 239 S N Say 1182 115m 1 and then 1181 gt 1 for all 239 gt m we can do this since we can arbitrarily index the conjugacy classes Must show m gt 1 N For some a gt 07 11G 19 m Z 1181 liGliZGQi im1 For this last number7 we know the numerator s a power of p other than 17 and the quo tient is an integer7 hence this term E 0 mod p This says m E 0 mod p7 which says m 2 p Then m 2 p gt1 implies m gt 17 QED Example of a group with a trivial center D67 which is isomorphic to 3 10 minutes left7 new thing Recall7 de nition a subgroup N of a group G is normal if sz l NVx E G Equiv alently7 we de ned the normalizer subgroup N0H G Exercise Lemma let h G a H be a homomorphism of groups Then kerH x E Glhw 611 is a normal subgroup Proof comes out of the de nition Given any x E G we need to show that skerHz 1 kerH That is7 we must show that each is a subset of the other But the reverse inclusion is equivalent to the forward inclusion zkerh 1 C kerh kerh C zkerh 1 gt zkerh 1 C kerh So zkerhx 1 E kerh Vm E G ltgt myz l E kerh Vm E G Vy E kerh Since h is a homomorphism of groups7 we have mm hltxgthltygthltxr1 hltzgthltzgt71 5H Hence myz l E kerh QED

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