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Complex Function Theory I

by: Mary Veum

Complex Function Theory I MATH 5120

Mary Veum
GPA 3.97


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This 12 page Class Notes was uploaded by Mary Veum on Thursday September 17, 2015. The Class Notes belongs to MATH 5120 at University of Connecticut taught by Staff in Fall. Since its upload, it has received 37 views. For similar materials see /class/205830/math-5120-university-of-connecticut in Mathematics (M) at University of Connecticut.


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Date Created: 09/17/15
Chapter 3 Quotient Groups and Automorphisms 31 De nitions and Examples De nition 311 Q5 G a H is a homomorphism Q5 1a is called a ber of Q5 over 1 De nition 312 The kernel of Q5 is kerqb g E Glow 1H Proposition 313 kerqb S G and MG 3 H This proposition is fairly trivial De nition 314 The Quotient Group or factor group of K kerqb in G is GK Xala 6 MG 3 H Where Xa is the bre of a E Multiplication is de ned for all a b 6 MG Xa 39 X12 Xab A quick proof that this is indeed a group 1 Assoiatiuity XaXch XaXbc Xabc Xabc X11ch XaXbXC 2 X1 ker Q5 is the identity 20 Quotient Groups and Automorphisms De nition 315 Consider the projection 7r G a GK de ned by Max XMQ for all Oz 6 G Proposition 316 7r is a homomorphism 065 X ltam X ltagt ltagt X ltagtX ltagt 00745 De nition 317 We de ne Q5 GK a H by Xa 1 Q5 is injective because Q5 is well de ned and Xa are the bers of Q5 over 1 Q5 is also quite clearly a homomorphism and of course Q5 is surjeetive onto imq imq S H This proves that GK N imq Also we can see that g5 obo Example 311 Let G Z and H ZnZ and de ne Q5 Z gt ZnZ by Ma 90 a E Z Assume 0 S a S n 1 Xa Q5 1690 m E Zl cm fl m E lem a 1 mEZlmNa modm a kew a5 1W Q5 10 0 Hz 32 The Quotient Group Proposition 321 Let Q5 G a H be a homomorphism K kerq If X E GK then X 1K for some a E X This means X uKlkeK KulkeK 33 THE HOLDER PROGRAM AND SOLVABILITY 21 Proof let X a Xa for some a E G Fix some u E X If as is another element of the ber then q5xu 1 1 aa 1 1 Thus you 1 h E kerqb But you 1u as E Ku So X Q Ku The other direction q5u 1x 1 implies that u 1x E ker Q5 Thus as E uKVx E X So X Q uK If uh E uK then q5ul a So uhEuK Q5 1a VuhEuK So uKQX This implies uK X for the other side hu 6 Ku h a a therefore hu 6 X so Ku Q X So Ku XandX Ku uK D De nition 322 Let H S G G a group G E G Then gH ghlh E H is a left coset of H in H Likewise Hg hglh E H is called a right coset of H in G Any element of a coset is called a representative of the coset because it can represent the coset Theorem 323 uKuK uuK for all uu E G or X uX v X ltugt ltvgtX ltuvgt Not thatfor ul 6 uK and 121 E uK uKuK ulKu1K u1u1K Sometimes we will write G N G and uK u because by hand it is quicker Theorem 324 The following are equivalent 1 GN is a group the multipication of cosets is well de ned 2 gNg 1 n or gN Ng so either N is closed under conjugation or equivalently left and right cosets are equal 3 N is the kernel of a homomorphism 33 The Holder Program and Solvability Theorem 331 The Holder Program To classify all simple groups the strategy consists of two parts 1 Classify all Finite Simple groups the periodic table of nite simple groups 22 Quotient Groups and Automorphisms 2 Final all the ways to put them all together De nition 332 A group G is called sovable Soluable if there is a tower 1 N0 N1 N2Elquot Nn G where NHlNi is abelian There are several alternate de nition of Sovlability of a group all of which are equivalent See the exercises Example 331 D4 is solvable 1 lt1 r lt1 Dn Exercise The Borel subgroup of GLnGF is solvable Proposition 333 Assume N 31 G then G is solvable if and only ifN and GN are solvable Proof 3 Because G is solvable we know there is a tower 1 GOlt101lt102lt1lt10n 0 Such that GHlGi is abelian For N lt1 G we construct a tower 1 N0lt1N1lt1lt1Nn N Where N G N N Now we see that N lt1 N241 Take n 6 Ni and g E N241 Thus n E G and n E N and g E G241 and N So gng 1 E G because G lt1 G241 and gng 1 E N because N is closed So gng 1 6 Ni so Ni lt1 G We see that NHlNi is abelian because L NHlN gt GHlGi is an injective homomorphism de ned as follows LaNi aGi This is well de ned because if aNi bNi then 1 lm for some n 6 Ni E It is a homomorphism because LaNibNi LabNi abGi aGibGi LaNiLbNi It is injective because if LaNi LbNi then aGi bGi so a by for someg E Gi Remembering ab E N we seeg b 1a E N so g E Gl N Ni So because NHlNi N lml S GHlGi it is abelian 1 34 EXERCISES 23 34 Exercises Exercise 311 94 Let Q5 G a H be a homomorphism and let E S H Prove that Q5 1E S G Also Show if Elt1 H Q5 1E lt1 G Proof Q5 1E is non empty 1h 6 E because E S H So q51g 1h 6 E thus 1g 6 Q5 Q5 1E is closed Let ab 6 Q5 ab q5aq5b E E So ab 6 Q5 Irwerses Let a 6 Q5 1E So q5a E E so a 1 q5a 1 6 Q5 Normal Now assume Elt1 H Let a 6 Q5 1E and g E G q5gag 1 ggwwo 1 ltggt ltagt ltggt 1 e E so gay 1 e a5 W is normal in D Exercise 3122 1 Prove that if H and K are normal subgroups of a group G then their intersection H N K is also a normal subgroup of 2 Prove that the intersection of an arbitrary nonempty collection of nor mal subgroups of a group is a normal subgroup do not assume the collection is countable Proof LetaEH KandgEG aEHsogagleHandaEKso gag 16K Sogag 1 EH K SoH KisnormalinG The above arguement abstracts easily to a collection of sets D Exercise 3131 94 Let N lt1 H S G show H S NgN People say this means that NgN is the largest subgroup of G in which N is normal Proof We already know that H is non empty and closed under multiplication and inverses all that we need to show is H Q NgN This follows from the fact that Nlt1H so for any h E H hNh 1 N and so it E NgN So if N is normal in any subgroup of G this subgroup of G is contained in the normalizer of N Making the normalizer the biggest in some sense subgroup of G in which N is normal D Exercise 3135 Prove that SLnF lt1 GLnl4 and describe the isomor phism type of the quotient group Proof Remember det GLnl4 gt IV is a homomorphism and the kernel is the set of martices with determinant 1 which is exactly SLnlF so SLnlF lt1 GLnlF We know that GLnlFSLnlF N lFX imdet D 24 Quotient Groups and Automorphisms Exercise 3140 94 Let N 31 G be a group Prove that xN commutes with yN in GN if and only if the commutator of x and y x 1y 19634 E N Proof Assume x 1y 19634 E N Then xyN ny Consider yx 1y 19634 acyeny Because Joy is also in xyN xyN and yacN are not disjoint and are thus the same D Exercise 3141 Let G be agroup prove that G x 1y 19531195 y E G is a normal subgroup of G and that GN is abelian N is called the com mutator subgroup Proof It is not the case that all the product of any two commutators need be a commutator However if you can prove that the generating set of a subgroup is closed under conjugation then the subgroup is closed under conjugation To see this we let 90 111239 an be an arbitrary product of elements of a generating set that is closed under conjugation in G Then 9x9 1 ga1a2ang 1 may 19a29 1gany 1 which is also an arbitrary product of generators So it is suf cient to show that the conjugate of a commutator is a commutator ie gacyg 1 is a commutator We get this by 91962119 1 9x 12 19629 1 9x 19 19y 19 19969 1929 1 9x9 1 19y9 1 19969 1gyg 1 9x9 17929 11 Which proves that G 31 G To show that GG is abelian we fall back on the previous exercise which shows that xG commutes with yG if and only if 9634 6 G But since G contains all commutators all cosets commute and G G is abelian 1 34 EXERCISES 25 Exercise 328 Let H and K be nite subgroups of G whose orders are relatively prime then H N K 1 Proof Let x E H and x E K Then divides both and 1K1 so 1 and thus 90 1 D Exercise 3211 Let H S K S G Prove that lG Hl lG KHK Proof Because cosets partition a group We see that G U MK ieI For some I such that 13161 are unique representatives of the cosets of K in G We see that I indexes the cosets of K in G so lIl lG Kl Likewise we can write K U ij jeJ Where J indexes the cosets of H in K So 1J1 lK This gives us 0 U 11ij ijeIgtltJ So we see that I X J indexes the cosets of H in G Thus lGHl lIgtlt Jl lIHJl lGKHKHl Which is our proof D Exercise 3212 Let H S G show that the map as 1 x 1 send each left coset of H in G onto a right coset of H and gives a bijection between the set of left cosets and right cosets of H in G Thus there are the same number of right cosets and left cosets Proof Let f G a G be de ned x 1 Vac E X We shall show that the image of the left coset gH fgH Hg 1 for any xed g E G LetghEngorsornehEH Thenh1EHsogh1 h1g16 Hg 1 And fgH Q Hg 1 Now we take an arbitrary kg 1 for some 16 E H Then 16 1 E H so gk 1 EgH and so fgk 1 gk 1 1 kg 1 E 26 Quotient Groups and Automorphisms So we have that the image of f is well de ned from the set of left cosets of H in G to the right cosets which maps gH H Hg 1 Using a similar argument above the image of f maps Hg 1 g 1H of right cosets to left cosets which is the inverse of the above map Showing that this map is bejective Exercise 3216 Use Lagrange s Theorem in the multiplicative ZpZgtlt to prove Fermat s little lemma If p is a prime then up N a mod p Proof First we know that lZpZgtltl q5p p 1 where Q5 is the Euler Phi Function We consider a 1 E ZleX noting that a E a and up 6 up Using Lagrange s theorem up 1 1 We get up cap 1 a SoapeasoapNa modp D Exercise 3222 Use Lagrange s Theorem to and the group ZrtZgtlt to prove Euler s Theorem aw N 1 mod rt for every a relatively prime to rt Proof Because art 1 we can consider a E ZTLZX By Lagrange s theorem aw 1 so aw E 1 so aw N 1 mod rt Exercise 334 Let A 31 G and B 31 D show that A X B 31 G X D and that G X DA X B N CA gtlt DB Proof Let ab E A X B and 9634 6 G X D Then xyabacy 1 rant 1yby 1 E A X B because Hoax 1 E A and yby 1 E B So A X B 31 C X D De ne Q5 Ggtlt DA X B gt GA gtlt DB be de ned by q5c dA X B bA CD This Q5 is a homomorphism and ker Q5 A X B which is trivial so it is an isomorphism D Exercise 337 and Let M and N be normal subgroups of G such that G MN Prove that GM N N gtlt GN Proof De ne Q5 G a gtlt GN such that a aMaN Q5 is an onto homomorphism with kernel equal to M Q N 34 EXERCISES 27 Homomorphic ab abMabN aMbMaNbN aMaNaNbN a bf0r all ab E G Surjective Let xMyN E gtlt GN Thus Ly E G Since G MN 90 mmnm and y myny for some mammy E M and nmny E N This means that 96M nmmmM an and yN Ny Nnymy Nmy myN So we consider nmmy nmmyM nmmyN an Nnmmy So Q5 is onto Kemb M Q N x6ker gt MN MN xEMandeN gtxEM N So we can invoke the rst homomorphism theorem to show Gkergb GM N N gtlt GN D Exercise 341 Show G 1 is a simple abelian group gt G N ZpZ for some prime p Proof For an abelian group G any N S G is normal in G So G being simple implies that G has no non trivial subgroups Let us take some a 1 E G a S G but since G has no non trivial subgroups this means that a G This implies not only that G is cyclic but it is generated by any of its elements 28 Quotient Groups and Automorphisms We now know that G must be nite because Z the only in nite cyclic group up to isomorphism is not simple So G must be nite we shall say 0 n gt 1 By the fundamental theorem of arithmetic there is pln By Cauchy s theorem this means there is some H S G such that lH l p Again since G is simple H G so lGl p We now have that G is a cyclic group of order p or G N ZpZ 1 Exercise 344 Show that if G is abelian and nHGl then there is so H S G such that n Proof We proceed by induction on the order of G The result is trivial when lGl 1 because this implies G We now use strong induction and claim that the result is true for all orders less than k Let lGl k and let n E Z be such that nlk so 16 m for some i E Zf By the fundamental theorem of arithmetic there is some p such that pln So n jp and k ijp for some j E Z3 By Cauchy s theorem we have there is some K S G such that lKl p Because G is abelian we know K 31 G so we can consider G K Observ ing that lGKl kp ij we can use our induction hypothesis to nd H S GK such that H j Finally consider H 7r 1H where 7r G a GK is the projection We see that H S G and pj n D Exercise 345 Show that subgroups of and quotient groups of a solvable group are solvable Proof Subgroups1 Let H be a subgroup of G we know G is solvable so there exists 1 G0lt1G1lt1ltIGT 0 Where GladG is abelian We let HI G N H we get the sequence 1 H0lt1H1lt1lt1HT H W pages 145 146 Note that the isomorphism theories he cites are under different names than he ones in Dummit and Foote and that there is o v in solvable 34 EXERCISES 29 Now we show that N lt1 N241 Take n 6 Ni and g E N241 Thus n E G and n E N and g E G241 and N So gng 1 E G because G lt1 G241 and gng 1 E N because N is closed So gng 1 6 Ni so Ni lt1 G We can show that the quotients are abelien because 0H1 m H 0H1 m H N 0mm m H H GmH Gm0mH GI The Last part is by the Diamond lsomorphism Theorem Quotien Groups Let Nlt1G then maintaining the G from above we establish the tower 1 NN CONNltIGlNNlt1ltIGTNN GNN GN We know N lt1 GiN because N S GiN S G and N lt1 G implies that N is normal in all intermediate subgroups We know GiNN lt1 G11NN because G lt1 G241 and GiNN 7rGI where 7r is the projection map from G to the cosets of N 7r is a homomorphism into G N so it is a homomorphism into G 11NN and the image of a normal subset is normal We see GHlNN N GMN GiNN GiN By the Third lsomorphism Theorem Now by the Diamond and Third lso morphism Theorems we see GI1N GI1GiN N GI1 N GPA0239 GiN GiN G241 GiN GHl GiNGi39 The last item is a quotient group of the abelian group GHlGi and so it is abelian itself So our tower is a solvable one and G N is solvable D Exercise 348 Let G be a nite group Show the following are equivalent It is apparent immediately that iiiiiiv implies igtii I m going to try to explain the proof taken from Lang s book We shall rst use strong induction on the order of an abelian group A The result is obvious for lAl 2 so assume lAl gt 2 then we can take 90 1 E A 30 Quotient Groups and Automorphisms and consider the cyclic group which is normal in A lA a l lt lAl so we can use the result and nd a cyclic tower 0 Ho ltxgt g H1ltxgt s g H ltxgt A ltxgt So we can nd a cyclic tower for A momma4H A Which shows what we wanted How does this help us with G Say we have the solvable tower 0 00301 quot39 Gr G We can apply our result to GHlGi to nd a cyclic tower between 7 G and GHl which gives us the tower we are looking for ii gtiii If we consider an arbitrary abelian group its composition fac tors must be abelian subgroups and quotient groups of abelian groups are abelian The only simple abelian groups are ZpZ where p is prime Through a similar process as the previous exercise we can construct a tower of these prime order simple cyclic groups iii gt iv Let M0 be the minimal normal subgroup of G We know that M0 is also solvable so by using the tower in iii we can nd Nlt1M0 such that M0 N is prime however this also means that MON is cyclic and abelian This means by exercise 3140 that all of the commutators of M0 are in N Because gNg 1 lt1M0 is of prime index the same must be true However this means that the commutators of M0 are contained in the intersection of the G conjugates of N which is a normal subgroup of G strictly contained inMO so by minimality of M0 all the commutators must be 1 So M0 is abelian We can apply the same concept to nd MlMO lt1GM0 which is solvable Here M1 31 G and MlMO is abelian So we construct a tower 0lt1M0lt1M1lt1lt1M5 G which satis es the given criteria


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