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# Linear Algebra MATH 214

GPA 3.7

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This 4 page Class Notes was uploaded by Maximilian Turcotte on Monday October 5, 2015. The Class Notes belongs to MATH 214 at Colgate University taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/219588/math-214-colgate-university in Mathematics (M) at Colgate University.

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Date Created: 10/05/15

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mu um 105 mlng am my Mam lass l ls alsy m We um a dawn a b aquot A mlng dugamquot m subm lp mm mm um mm wme hue mm 13 mm le Ma 40mm the luva 5 mlng Thetad muodlms the 4 mm mamas whae heal W we by Wu pm on m4 m l S 40 mm butthetmmlds um mum nunlg mm W 5 the spamquot mm 1310 m w Ammtadnckmse ls mm L m m 7L dugam W m the rum ls the mlng Mme l gt I saw thls gxoup hlst ln the context ol the algebla ol quatelnlonsquot ln the mole common notatlon 1 lol I 1 lol I 7 lol K and lo lol L The algebla ol quatelnlons ls the set ol expxesslons ebee7 dc whele obed 7 lR It ls an extenslon om ovel lR Oul text would deal Wlth lt as a subalgebla ol szz ovel W The symmetllc gxoup on n lettels 52 l e the set ol one7to7one lunctlons llom 12 7 p undel composltlon has a sungoup the set ol all elements ol 5n lol Whl n 7 n7 l e the lu ctlons that n t move to allh cleal that thls sungoup behaves exacth llkequot5m1 ndt nothl speclal hele ab tth t 1 the su et n F lany s nd subset Y ol t e mmetllcgl up 5 h asubgloup con g s t ol all elements Whl 7 yl vely ym Y whlch l e behaves exacth lllcequot 5x 7 Y The subset ol all elements ol 5x lol whlch My 7 Y every 3 ln Y and c 7 x 7 Y lol every c 7 x 7 Y ls anothel sungoup ol 5x ll Y ls ln nlte then a lunctlon may ta e ln ke Y lnto Y Ebcample The add onequot lunctlon ap 7 cc 1 ls an element ol 51 lol whlch subset ol 51 conslstlng ol lunctlons lol whlch Z g 1 ls not a sungoup ol 51 ause lt ls n t c osed undex lnvezses Let a be any gxoup and 9 be a xed element ol 6 Then Za 9 7 a gas 7 be the set ol all elements that commute vvlthquot a9 ls easlh checked to be a sungoup ol 6 called the eeoclotmt ol 9 Agaln thele ls nothlng speclal about a slngleelement set no Fol a subset x ol 6 the centlallzel ol 0 z x 7 g 7 a gas 7 9 w 7 ol 6 We could make thls a cozollary ol the lesult below that an lntelsectlon ol a laml ol subgzoups ls agaln a subgloup because zx 7 Zw a9 7 x In paltlculal the centlallzel ol 6 ltsell 2a the set ol all elements ol 6 that commute Wlth every element ol 6 ls a subgloup called the canth ol 6 Some examples ol centels and centlallzels bellan then ol coulse 26 7 genelally ll 9 7 26 then Za 7 a t Y L K I t commute Wlth evelythlng But z1 7 none ol the othel loul elements ol Q3 do a and lol each element 9 ol 6 Zw a Mole 71 I the bowels ol J commute Wlth I but I claim that ZGL2 R a1 a 6 IR H 0 the set of nonzero scalar matrices It is easy to see that these matrices commute with every element of GL2 IR so we need 2 gt e ZGL2 IR then lt3 gtlt32gtlt33gtlt gt 33gtltSf gtv socbandad Also 10 abiab 10 j a b iabb 11 ba 7 ba 11 baab 7 baa so a a b and hence I 0 S0 A a for some nonzero scalar a to see that no other elements do so Suppose A lt Z We can list the 3 6 elements ong and one ofthem is f 1 H 2 2 H 3 3 H 1 We can check that f2 1 H 3 2 H 13 H 2 and that f3 e the identity function The other three elements of 3 reverse two of the numbers 1 2 3 and leave the third xed and we can check that they do not commute with 1 For example letting g 1 H 2 2 H 1 3 H 3 fog1gt gt32gt gt23gt gt1 but gof1gt gt12gt gt33gt gt2 SO Zltfgt ltfgt And ZSs 6 Prop Let HA E A be a family of subgroups of the group G Then the subset H H E A of G consisting of the elements that are in every one of the H s is also a subgroup of G Pf Because the identity is in every one of the H s it is in H so H is not empty To see that H is closed under the operation take m y in H then because each H is a subgroup and contains m y it also contains their product my so their product is also in H To see that that H is closed under inverses take z in H then because z is in each H so is z l so m 1 E The text shows that the union of two subgroups is never a subgroup unless it is one of them ie unless one is contained in the other The union of more subgroups may be a subgroup but it probably isn t We can describe the subgroups of a cyclic group Prop Let G be a cyclic group i Every subgroup of G is cyclic ii If G is in nite cyclic then the subgroups of G are e G 2 3 all distinct iii If G is nite cyclic of order n then for each divisor d of n G has exactly one subgroup of order 1 namely lm and it has no other subgroups Pf The trivial subgroup e is cyclic generated by 5 so take a nontrivial subgroup H of G Then H contains some positive power of the generator m suppose the smallest such power is M We want to show that H Because zk E H we have Q H For the reverse inclusion take any element mm of H and long divide m by k say m dk r where d r E Z and 0 g r lt k Then zm dk zmmk d E H because H is closed under the operation and inverses but k was chosen so that zk was the smallest positive power of z in H so r can t be positive ie r 0 Thus mm zkd E M and so H Q ii Clearly the sets of powers of zk and m k are the same set so the given list includes all the subgroups of G so we only have to show that they are distinct if G is in nite cyclic The text leaves this as an exercise for the reader so I will too iii Now we are supposing that G is nite of order 71 From the proof given in i any nontrivial subgroup H of G has the form where k is the smallest positive integer for which zk E H We want to show that this k is a divisor of n We know z e E H and long dividing n by k shows just as in i that kln say 71 kd We proved in Section 4 that 0zk n gcdn k nk so is a subgroup of G of order Thus every subgroup of G has order a divisor nk d of n and is generated by zk mnd Conversely if d is a divisor of n then we also proved in Section 4 that zn i is an element of order n gcdn 1 so lm is a subgroup of order Cor In a nite cyclic group G of order n ngd gtkgt In particular z ms iff gcdn r gcdn s Pf For the rst equality we need to show that each of zk and mng gtk is a power of the other Because k is a multiple of gcdn k zk is a power of ngd gtk and because gcdn k nr ks for some r s in Z we have mng gtk zn7zkS Thus ngd gtkgt It follows that if gcdnr gcdn s then x Conversely if z ms then mng gtTgt ngd gt9gt and because the subgroups are equal and the exponents are factors of n the exponents gcdn r and gcdn r are also equal The text also includes the following useful fact If S is a nite nonempty subset of a group G and S is closed under the operation then S is also closed under inverses and hence is a subgroup of G The proof is simple Let x E S then because S is nite and closed under the operation the powers of z cannot all be different say mp zq where p lt q We get zq p 5 so x has nite order and its inverse is a positive power of it so m 1 E S

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