Mth 94 Professor: Pri Brtlett Leture 8: Astrt Alger Week 8 UCSB 2015 This is the eighth week of the Mthemtis Sujet Test GRE prep ourse; here, we run very rough-n-tumle review of strt lger! As lwys, this fiel is muh igger thn one lss; oringly, we fous our ttention on key efinitions n results. 1 Groups: Definitions n Theorems Definition. A group is set G long with some opertion tht tkes in two elements n outputs nother element of our group, suh tht we stisfy the following properties: Ientity: there is unique ientity element e G suh tht for ny other g G, we hve e g = g e = g. Inverses: for ny g G, there is unique g 1 suh tht g g 1 = g 1 g = e. Assoitivity: for ny three,, G, ( ) = ( ). Definition. A sugroup H of group G, is ny suset H of G suh tht H is lso group with respet to the opertion. Definition. A group G, is lle elin, or ommuttive, if it stisfies the following itionl property: Commuttivity: for ny, G, =. There re mny ifferent exmples of groups: Exmple. 1. The rel numers with respet to ition, whih we enote s R, +, is group: it hs the ientity 0, ny element x hs n inverse x, n it stisfies ssoitivity. 2. Conversely, the rel numers with respet to multiplition, whih we enote s R,, is not group: the element 0 R hs no inverse, s there is nothing we n multiply 0 y to get to 1! 3. The nonzero rel numers with respet to multiplition, whih we enote s R,, is group! The ientity in this group is 1, every element x hs n inverse 1/x suh tht x (1/x) = 1, n this group stisfies ssoitivity. 4. The integers with respet to ition, Z, + form group! 5. The integers with respet to multiplition, Z, o not form group: for exmple, there is no integer we n multiply 2 y to get to 1. 1
6. The nturl numers N re not group with respet to either ition or multiplition. For exmple: in ition, there is no element 1 N tht we n to 1 to get to 0, n in multiplition there is no nturl numer we n multiply 2 y to get to 1. 7. GL n (R), the olletion of ll n n invertile rel-vlue mtries, is group uner the opertion of mtrix multiplition. Notie tht this group is n exmple of[ nonelin group, s there re mny mtries for whih AB BA: onsier ] 0 1 0 0 [ ] [ ] [ ] [ ] [ ] 1 0 0 0 1 0 0 1 0 1 = versus =. 0 0 0 0 0 0 0 0 0 0 8. SL n (R), the olletion of ll n n invertile rel-vlue mtries with eterminnt 1, is lso group uner the opertion of mtrix multiplition; this is euse the property of eing eterminnt 1 is preserve uner tking inverses n multiplition for mtries. 9. The integers mo n, Z/nZ is group with respet to ition! As reminer, the ojet Z/nZ, +, is efine s follows: Your set is the numers {0, 1, 2,... n 1}. Your ition opertion is the opertion ition mo n, efine s follows: we sy tht + mo n if the two integers + n iffer y multiple of n. For exmple, suppose tht n = 3. Then 1 + 1 2 mo 3, n 2 + 2 1 mo 3. Similrly, our multiplition opertion is the opertion multiplition mo n, written mo n, n hols whenever + n iffer y multiple of n. For exmple, if n = 7, then 2 3 6 mo 7, 4 4 2 mo 7, n 6 4 3 mo 7. 10. (Z/pZ) = {1,... p 1} is ommuttive group with respet to the opertion of multiplition mo p, if n only if p is prime. Seeing this is not too iffiult, n is useful thing to o to remin ourselves out how moulr rithmeti works: It is esy to see tht (Z/pZ) stisfies ssoitivity, ientity n ommuttivity, simply euse these properties re inherite from the integers Z: i.e. if =, then surely = mo p, euse equlity implies equivlene mo p! Therefore, the only property we nee to hek is inverses. We first el with the se where p is not prime. Write p = mn for two positive integers m, n 1; notie tht euse oth of these vlues must e smller thn p if their prout is p, oth m n n live in the set {1,... p 1}. 2
Consier the element n. In prtiulr, notie tht for ny k, we hve kn x mo p kn x is multiple of p kn x is multiple of mn kn x is multiple of n x is multiple of n. (If none of the ove eutions mke sense, reson them out in your he!) Beuse of this, we n see tht n hs no inverse in (Z/pZ), s kn is only ongruent to multiples of n, n 1 is not multiple of n. The onverse showing tht if p is prime, (Z/pZ) hs inverses is little trikier. We o this s follows: first, we prove the following lim. Clim. For ny, {0,... p 1}, if 0 mo p, then t lest one of, re equl to 0. Proof. Tke ny, in {0,... p 1}. If one of, re equl to 0, then we know tht = 0 in the norml multiplying integers worl tht we ve live in our whole lives. In prtiulr, this mens tht 0 mo p s well. Now, suppose tht neither nor re equl to 0. Tke oth n. Rell, from gre shool, the onept of ftoriztion: Oservtion. Tke ny nonzero nturl numer n. We n write n s prout of prime numers n 1... n k ; we think of these prime numers n 1,... n k s the ftors of n. Furthermore, these ftors re unique, up to the orer we write them in: i.e. there is only one wy to write n s prout of prime numers, up to the orer in whih we write those primes. (For exmple: while you oul sy tht 60 n e ftore s oth 2 2 3 5 n s 3 2 5 2, those two ftoriztions re the sme if we on t re out the orer we write our numers in.) In the speil se where n = 1, we think of this s lrey ftore into the trivil prout of no prime numers. Tke, n write it s prout of prime numers 1... k. Do the sme for, n write it s prout of primes 1... m. Notie tht euse n re oth numers tht re stritly etween 0 n n 1, n nnot e one of these prime numers (euse positive multiples of n must e greter thn n!) In prtiulr, this tells us tht the numer on one hn n e written s the prout of primes 1... k 1... m, n on the other hn (euse ftoriztions into primes re unique, up to orering!) tht there is no n in the prime ftoriztion of. Conversely, for ny nturl numer k, the numer k n must hve ftor of n in its prime ftoriztion. This is euse if we ftor k into prime numers k 1... k j, we hve k n = k 1... k j n, whih is ftoriztion into prime numers n therefore (up to the orer we write our primes) is unique! 3
In prtiulr, this tells us tht for ny k, the quntities n k p re istint; one of them hs ftor of p, n the other oes not. Therefore, we hve shown tht if oth n re nonzero, then nnot e equl to multiple of p in other wors, is not ongruent to 0 moulo p! Therefore, the only wy to pik two, {0,... p 1} suh tht is ongruent to 0 moulo p is if t lest one of them is equl to 0, s lime. From here, the proof tht our group hs inverses is pretty strightforwr. Tke ny x (Z/pZ), n suppose for ontrition tht it i not hve ny inverses. Look t the multiplition tle for x in (Z/pZ) :... p 1 x???...? If x oesn t hve n inverse, then 1 oes not show up in the ove tle! The ove tle hs p slots, n if we re trying to fill it without using 1, we only hve p 1 vlues to put in this tle; therefore some vlue is repete! In other wors, there must e two istint vlues k < l with xl xk mo p. Consequently, we hve x(l k) 0 mo p, whih y our ove oservtion mens tht one of x, (l k) re equl to 0. But x is nonzero, s it s tully in (Z/pZ) : therefore, l k is equl to 0, i.e. l = k. But we si tht k, l re istint; so we hve ontrition! Therefore, every element x hs n inverse, s lime. 11. The symmetri group S n is the olletion of ll of the permuttions on the set {1,... n}, where our group opertion is omposition. In se you hven t seen this efore: A permuttion of set is just ijetive funtion on tht set. For exmple, one ijetion on the set {1, 2, 3} oul e the mp f tht sens 1 to 2, 2 to 1, n 3 to 3. One wy tht people often enote funtions n ijetions is vi rrow nottion: i.e. to esrie the mp f tht we gve ove, we oul write f : This, however, is not the most spe-frienly wy to write out permuttion. A muh more onense wy to write own permuttion is using something lle yle nottion. In prtiulr: suppose tht we wnt to enote the permuttion tht sens 1 2, 2 3,... n 1 n, n 1, n oes not hnge ny of the other elements (i.e. keeps them ll the sme.) In this se, we woul enote this permuttion using yle nottion s the permuttion ( 1 2 3... n ). 4
To illustrte this nottion, we esrie ll of the six possile permuttions on {1, 2, 3} using oth the rrow n the yle nottions: i : (23) : 1 2 3 (12) : (123) : 1 2 3 (13) : (132) : 1 2 3 The symmetri group hs severl useful properties. Two notle ones re the following: Clim. We n write ny σ S n s prout of trnspositions 1. Clim. For ny finite group G, there is some n suh tht G is sugroup of S n. 12. The iherl group of orer 2n, enote D 2n, is onstrute s follows: Consier regulr n-gon. There re numer of geometri trnsformtions, or similrities, tht we n pply tht sen this n-gon to itself without strething or tering the shpe: i.e. there re severl rottions n refletions tht when pplie to n-gon o not hnge the n-gon. For exmple, given squre, we n rotte the plne y 0, 90, 180, or 270, or flip over one of the horizontl, vertil, top-left/ottomright, or the top-right/ottom-left xes: (rotteryr0 ) (rotteryr90 ) (rotteryr180 ) (rotteryr270 ) (fliproverrhorizontl) (fliproverrvertil) (fliproverrul/drrigonl) (fliproverrur/dlrigonl) 1 A permuttion σ S n is lle trnsposition if we n write σ = (), for two istint vlues, {1,... n}. 5
Given two suh trnsformtions f, g, we n ompose them to get new trnsformtion f g. Notie tht euse these two trnsformtions eh iniviully sen the n-gon to itself, their omposition lso sens the n-gon to itself! Therefore omposition is well-efine opertion tht we n use to omine two trnsformtions. (rotte y 90 ) (flip over vertil) = (flip over UR/DL igonl) This is group! Now tht we hve some exmples of groups own, we list some useful onepts n efinitions for stuying groups: Definition. Tke ny two groups G,, H,, n ny mp ϕ : G H. We sy tht ϕ is group isomorphism if it stisfies the following two properties: 1. Preserves size: ϕ is ijetion 2. 2. Preserves struture: ϕ, in sense, sens to. To esrie this formlly, we sy the following: g 1, g 2 G, ϕ(g 1 g 2 ) = ϕ(g 1 ) ϕ(g 2 ). This property preserves struture in the following sense: suppose tht we hve two elements we wnt to multiply together in H. Beuse ϕ is ijetion, we n write these two elements s ϕ(g 1 ), ϕ(g 2 ). Our property sys tht ϕ(g 1 g 2 ) = ϕ(g 1 ) ϕ(g 2 ): in other wors, if we wnt to multiply our two elements in H together, we n o so using either the G-opertion y lulting ϕ(g 1 g 2 ), or y using the H-opertion y lulting ϕ(g 1 ) ϕ(g 2 ). Similrly, if we wnt to multiply ny two elements g 1, g 2 in G together, we n see tht g 1 g 2 = ϕ 1 (ϕ(g 1 g 2 )) = ϕ 1 (ϕ(g 1 ) ϕ(g 2 )). So, gin, we n multiply elements using either G or H s opertion! To hoose whih opertion we use, we just nee to pply ϕ or ϕ 1 s pproprite to get to the esire set, n perform our lultions there. 2 Notie tht this mens tht there is n inverse mp ϕ 1 : H G, efine y ϕ 1 (h) = the unique g G suh tht ϕ(g) = h. 6
Definition. Tke ny two groups G,, H,, n ny mp ϕ : G H. We sy tht ϕ is group homomorphism if it stisfies the preserves struture property ove. Theorem. If G, H re groups n ϕ : G H is homomorphism, then ker(ϕ) = {g G ϕ(g) = i} is sugroup of G, n for ny sugroup S of G, ϕ(s) = {ϕ(s) s S} is sugroup of H. Definition. A sugroup H of group G is lle norml if for ny g G, the left n right osets 3 gh, Hg re equl. We write H G to enote this property. Theorem. Suppose G is group n H is norml sugroup. Define the set G/H to e the olletion of ll of the istint left osets gh of H in G. This set forms something lle the quotient group of G y H, if we efine g 1 H g 2 H = (g 1 g 2 )H. This is useful onstrution, n omes up ll the time: for exmple, you n think of Z/nZ s quotient group, where G is Z n H = nz = {n k k Z}. Definition. Tke ny group G, of orer n: tht is, ny group G onsisting of n istint elements. We n rete group tle orresponing to G s follows: Tke ny orering r 1,... r n of the n elements of G: we use these elements to lel our rows. Tke ny other orering 1,... n of the n elements of G: we use these elements to lel our olumns. (This orering is usully the sme s tht for the rows, ut it oes not hve to e.) Using these two orerings, we rete n n rry, lle the group tle of G, s follows: in eh ell (i, j), we put the entry r i j. Theorem. Two groups G,, H, re isomorphi if n only if there is ijetion ϕ : G H suh tht when we pply ϕ to group tle of G, we get group tle of H. Theorem. (Cyley.) Let G, e finite group, n g G e ny element of G. Define the orer of g to e the smllest vlue of n suh tht g n = i. Then the orer of g lwys ivies the totl numer of elements in G, G. More generlly, suppose tht H is ny sugroup of G. Then H ivies G. This theorem hs useful speil se when we onsier the group (Z/pZ) : Theorem. Fermt s Little Theorem. Let p e prime numer. Tke ny 0 in Z/pZ. Then p 1 1 mo p. 3 The left oset gh of sugroup H y n element g is the set {g h h H}. Bsilly, it s H if you t on eh element y g. Right osets re the sme, ut with Hg inste. 7
Theorem. Any finite elin group G is isomorphi to iret sum 4 of groups of the form Z/p k j j Z. In other wors, for ny finite elin group G, we n fin primes p 1,... p l n nturl numers k 1,... k l suh tht G = Z p k 1 1 Z k p l. l Groups re not the only lgeri ojets people stuy: 2 Rings n Fiels: Definitions n Theorems Definition. A ring is set R long with two opertions +, so tht R, + is n elin group. R, stisfies the ssoitivity n ientity properties. R stisfies the istriutive property: i.e. for ny,, R, we hve ( + ) = +. The ientity for + is not the ientity for. Some people will enote ring with multiplitive ientity s ring with unity. I elieve it is slightly more stnr to ssume tht ll rings hve multiplitive ientities, n in the o instne tht you nee to refer to ring without multiplitive ientity s rng. Mny of the exmples we sw of groups re lso rings: Exmple. 1. The integers with respet to ition n multiplition form ring, s o the rtionls, rels, n omplex numer systems. 2. The Gussin integers Z[i], onsisting of the set of ll omplex numers { + i, Z, i = 1} form ring with respet to ition n multiplition. Z/nZ is ring for ny n. Definition. A integrl omin is ny ring R where the following property hols: For ny, R, if = 0, then t lest one of or is 0. Exmple. 1. The integers with respet to ition n multiplition form integrl omin, s o the rtionls, rels, n omplex numer systems. 4 A group G is lle the iret sum of two groups H 1, H 2 if the following properties hol: Both H 1, H 2 re norml sugroups of G. H 1 H 2 is the ientity; i.e. these two sugroups only overlp t the ientity element. Any element in G n e expresse s finite omintion of elements from H 1, H 2. We think of G = H 1 H 2. 8
2. Z/nZ is not n integrl omin for ny ompositite n: if we n write n = for two, < n, then we hve tht 0 mo n, while neither, re multiples of n (n thus not equivlent to 0). Definition. A fiel is ny ring R where R, is group. (By R, we men the set of ll elements in R other thn the itive ientity.) Exmple. 1. The integers with respet to ition n multiplition re not fiel. 2. The rtionls, rels, n omplex numer systems re fiels with respet to ition n multiplition! 3. Z/pZ is fiel with respet to ition n multiplition. There re mny mny theorems out rings n fiels; however, the GRE will not require you to know lmost ll of them. Inste, they mostly wnt you to e fmilir with wht they re, n how they re efine! To illustrte how the GRE tests you on these onepts, we run few prtie prolems here: 3 Smple GRE prolems Prolem. Consier the set G me out of the four omplex numers {1, 1, i, i}. This is group uner the opertion of omplex multiplition. Whih of the following three sttements re true? 1. The mp z z is homomorphism. 2. The mp z z 2 is homomorphism. 3. Every homomorphism G G is of the form z z k, for some k N. () 1 only. () 1 n 2 only. () 2 n 3 only. () 1 n 2 only. (e) 1, 2 n 3. Answer. We n nswer this prolem quikly y lssifying ll possile homomorphisms ϕ : G G. We n first notie tht we must sen 1 to 1 if we re homomorphism. To see this, notie tht ϕ(1) = ϕ(1 1) = ϕ(1) ϕ(1), n therefore y neling ϕ(1) on oth sies we hve 1 = ϕ(1). Now, onsier where to sen i. If we hve ϕ(i) = i, then we must hve ϕ( 1) = ϕ(i 2 ) = ϕ(i)ϕ(i) = i i = 1, n ϕ( i) = ϕ(i 3 ) = ϕ(i)ϕ(i)ϕ(i) = i 3 = i. So we re the ientity, n lso we re the mp z z 3. If we hve ϕ(i) = 1, then we must hve ϕ( 1) = ϕ(i 2 ) = ϕ(i)ϕ(i) = 1 1 = 1, n ϕ( i) = ϕ(i 3 ) = ϕ(i)ϕ(i)ϕ(i) = 1 3 = 1. So we re the mp tht sens everything to 1; lterntely, we re the mp z z 4. Similrly, if we hve ϕ(i) = 1, then we must hve ϕ( 1) = ϕ(i 2 ) = ϕ(i)ϕ(i) = 1 1 = 1, n ϕ( i) = ϕ(i 3 ) = ϕ(i)ϕ(i)ϕ(i) = ( 1) 3 = 1. So we re the mp z z 2. 9
The lst possiility is if we hve ϕ(i) = i, then we must hve ϕ( 1) = ϕ(i 2 ) = ϕ(i)ϕ(i) = i i = 1, n ϕ( i) = ϕ(i 3 ) = ϕ(i)ϕ(i)ϕ(i) = ( i) 3 = i. So we re the mp z z, or lterntely the mp z z 3. As result, ll of the lims 1,2,3 re ll true; so our nswer is e. Prolem. Let R e ring. Define right iel of R s ny suset U of R suh tht U is n itive sugroup of R. For ny r R, u U, we hve ur U. Suppose tht R only hs two istint right iels. Whih of the following properties must hol for R? 1. R ontins infinitely mny elements. 2. R is ommuttive. 3. R is ivision ring; tht is, every nonzero element in R hs multiplitive inverse. () 1 only. () 2 only. () 3 only. () 2 n 3 only. (e) 1, 2 n 3. Answer. We first notie tht ny ring lwys hs two iels, nmely {0} n R. The first property is eliminte y notiing tht R = Z/2Z is ring. Its only itive sugroups re {0} n R, so in prtiulr those re its only two iels, n this group is lerly finite. The seon property is eliminte y relling the quternions H, whih re nonommuttive ring! Finlly, we n verify tht the thir property must hol. To see this, tke ny, n onsier the set R = {r r R}. This is n itive sugroup, n lso right iel! Therefore it is either the ll-zero sugroup (only if = 0, s otherwise 1 = 0) or ll of R. But this mens tht there is some r R, s R suh tht rs = 1; i.e. rs = 1. So hs n inverse! This leves 3 s the only possiility. 10