MTH421-Lecture14.pdf

MTH4211521 Lecture 14 9121 20

Det Let G be a group H a subgroup For eachelement a EG the left coset oft associated with a is the set

at haha h EH's

In additive notation atte hath htt 9

we canthinkof a H as some sort of shift of H by a The right coset oft associated with a is defined as

Ha L ha HEH 9 Reward for each a aft may not equal Ha

EI Let G Zg H 244 Then the left Cosets are

0 184 10,2 4,69 I124 21,3 5,79 2 184 42.46,09 3 124 23 5,7 IE 4 1 44 24 6 0,29 51 24 45,8 1,33 6 t Za 26,0 2.49 7124 47 I 359

Note that 0124 2124 4 124 6424 70,246 It 24 3 24 5424 7 24 21,35,79

EI Let G Sss H La B I write down all the left cosees of H

Recall the elements of G ave

C 331 43 l 33 GD

L23 uz 1223 clad

f I 132 EI CB

LD H H B It L 233cL 31113 1235 42334 C12 H L 112 42311339 L R 113238 123 H 411237 11233433 11237 23 432 SHE 46132 13234339 2932 112J's B H L 1137 4374339 Lab 4379 14

Note CDH 4 B H uh 11376 IN 123H 4 237423 2 It C1323ft iz 113239

coincidence

Lemme Properties of Cosets Let G be a group and H a subgroup Let a b EG Then

I a C AH 2 aH H if and only if a EH 3 CabSH albH 4 AH bH if andonly if aEbH 5 Either aH bH or at Nbt _of 6 aH bH it andonly if a b EH 7 late4171 8 att Ha if andonly if H aHaY 9 att is a subgroup of G if andonlyif

a Etl

tf 1,43 Exercises 4 First suppose a Ebt Then a bhp for some h EH We show attebH and bH Eat Let X be any element of a H

Then A ah for some h E H Now f ah bheh bChih since it is closed hehEH so debt

Since a bhi b ahit Let y be anyelement of BH Then Y bh for some HEM Now

y bh ahi h aChi h Gatt where we used the fact that hith EH Since XYarearbitrary we have ate BHand BH Eat So at bt

5 Suppose att n BHI d Lets be an element in aHnbH then

I ah bhz for some hihaGH a bhahi Ebt

By 4 att BH

6 By 4 aH bH if and onl if aEbH Now a Ebt Eh EH a bh

7 HEH b a hs b la EH at BEM

7,8 Exercises

9 By 5 the distinct coset of H partition G Amongthem only It contains e and is a subgroup By 4 aH eH H if andonly if aEH

By Lemma itemsCD the distinct coset of H form a partition of G

100002 Ee Find the distinct co sets of H 41,3 5,9 159 in G U Zz L1,3 37,9 B 15,17 19,219

Weknow 1H 3 test 9H 45 H L1,3 5,9159

Now pick an element not in H find its cosee say 7 Ftl L 7 21,35 63 1059 mod22

47,21 13 19 179 BY the lemma

7 A 1311 1714 1914 21 HE471317,1948

So these two are the distinct coset

Thur71 Lagrange's thus Let G be a finite group and H a subgroupofG Then IHI divides G Morever the number of distinct Clefts assets of ft in Cr in 1611411 PI By Lemma eD ta EGe a Gat So alfgate G By E Va.BE G either at bat or aH n bit of So if a H Azt AmH are all the distinct Cosets then they must be pairwise disjointand their union is G So they form a partition of G By CF AiH1492141 1amHEL HM SO 1Gt 2192171 dueto the partition

f I

m CHI

SO m IGI th B implies Hel11611h11 and that distinct assets of Hin G e E

Reward the number of left Cosets of H in G is also called the index of Hin Gand is denotedby 1 Git

Y 1

Cord If G B a finite groupand H a subgroup then 1betel 161 1141

Core If G is a finite group the order lad of any element a must divide Gl PI sa is a cyclic subgroupof G oforder lat By thin F l lal must dicide G

EI U122 L I 3,5 7,9 13 15,17 19,219 WED1 10 the only possible

ordersof an element are 1 2,5 or lo For instance we saw earlier that 131 5

Cor3 Let p be a prime Let G be a group of order p Then G must be cyclic PI Let a be any nonidentity element Then lat By Cor 3 lal mustdivide p But the only divisors otp are 1 and P so at p So G La

Cort Let G be a group of finite order n Let a EG Then a ee PI Let metal By Cor 3 mln So ne m K for some integer K Now

an Cam keek e

Corte Fermat's little Theorems For every integer a andeveryprime p APmod p a modp

PI By the division algorithm a mptr where m r EE and o Er Ep 1 So amodp r By properties of modulo operation V xiyifxmodp y modp henxnmudp ynmodp.fr all n EET

So it suffices to prove that pPmed p r

consider the group VCR L1,2 n p 19

where the group operation is multiplication modulo p By Cor 4

getmod p I

By properties of modulo operation V a.bi.DE ifamodp bmodpand cmodp dmodp then acmadp abdmodp So rMmodp I made

rmodp rmodp rPmodp rmodp h EB