MAT 444 H Barclo Spring 004 Homwork 6 Solutions Sction 6 Lt H b a subgroup of a group G Thn H oprats on G by lft multiplication Dscrib th orbits for this opration Th orbits of G ar th right costs of H 4 Prov th Fixd Point Thorm () W must show that if G = p, p, a prim intgr, and S is a finit G-st such that p s, thn s S such that Gs = G If w show that thr is on orbit Os with O s = thn Gs = G by th counting formula Lt k b th numbr of distinct orbits of S undr th action of G Thn i i i k By th counting formula j /,, S = O O O O S i = k, thus r O i j = p whr O ri If all ri, thn p would divid S Thus, at last on r i = 0, which yilds an orbit of siz, say O s and s is a dsird fixd point 8 Dtrmin th Class Equation for ach of th following groups a) th quatrnion group a) Conjugacy classs: {}, { }, { i, }, { j, j}, { k, k} Class quation: 8 = + + + + b) th Klin four group
MAT 444 Barclo Homwork 6 Pag b) Klin group G is ablian, thus vry lmnt of G blongs to a conjugacy class mad of that lmnt only Class quation: 4 = + + + c) th dihdral group D c) s () d) D 6 d) s () ) D n ) G = Dn Cas : n is odd Conjugacy classs: Class quation: Cas : n is vn Conjugacy classs: Class quation: n n+ n {}, { x, x },, { x, x }, n { y, xy, x y,, x y} n = + + + + + n n factors n n n + n {}, { x, x },, { x, x }, { x }, n 3 n { y, x y,, x y},{ xy x y,, x y} n n + + + + + + + n factors f) th group of uppr triangular matrics in GL( F 3)
MAT 444 Barclo Homwork 6 Pag 3 f) Conjugacy classs: SL ( F ) 0 0,,,,, 0 0 0 0 0 0 0 0,,,,, 0 0 0 0 0 0 class quation: + + + + 3 + 3 g) 3 g) class quation: + + 4 + 4 + 4 + 4 + 6 6 Prov that no group of ordr Sction 6 p, whr p is prim and >, is simpl Lt G b a group of ordr p, p prim and > By proposition () ZG ( ) > But Z( G) G G, so if Z(G) is not all of G thn G is not simpl If Z( G) = G, this mans that G is ablian But G has at last an lmnt x of ordr p, hnc x < G and x G, sinc G is ablian Sction 63 6 Considr th opration of lft multiplication by G on th st of its substs Lt U b a subst whos orbit {gu} partitions G Lt H b th uniqu subst in this orbit which contains Prov that H is a subgroup of G and that th sts gu ar its lft costs W hav that i gu partitions G, so blongs to on subst, say H = gu But thn i gh also partitions G Lt G H = { g G gh = H } b th stabilizr of H, w show that GH = H, hnc H is a subgroup of G
MAT 444 Barclo Homwork 6 Pag 4 i) If h H thn h = h hh, which mans that h H hh H = hh h Gh; thus, H GH ii) Lt s GH thn sh = H, and sinc H, s sh = H, hnc GH H 3 a) Lt H b a normal subgroup of G of ordr Prov that H is in th cntr of G a) Lt H = {, x} G axa H, a G But axa sincif axa = x = Thus Sinc clarly Z( G) thn H Z( G) axa x ax xa x Z G = = ( ) b) Lt H b a normal subgroup of prim ordr p in a finit group G Suppos that p is th smallst prim dividing G Prov that H is in th cntr Z(G) b) Lt H G with H = p Lt x and x H such that H Z( G) W show that for all g G th st Cx = { gxg g G} =, thus sinc x Cx w hav gxg = x g G gx = xg g G W know that C x divids th ordr of G Morovr, sinc H G, gxg H g G Cx H Hnc Cx H = p, sinc p is th smallst prim that divids G Cx = p or If Cx = p w hav that Cx = H which mans g st gxg =, i x =, contradicting our assumption Thus C x = and if x H and x x Z( G) Clarly, Z( G) as wll H Z( G) Sction 64 a 4 Prov that th st of matrics c whr ac, F and c=,, 4 forms a group of th typ prsntd in (49b) (and that thrfor such a group xists)
MAT 444 Barclo Homwork 6 Pag a Lt G = a, c, c =,, 4 F No that H = {,, 4} is a cyclic subgroup of th multiplicativ group F {0} 0 Clarly G 0, thus thr is an idntify lmnt in G a ac a Lt G, thn =, but sinc H is a subgroup c a G a a Lt A = and B = 0 b with c, b H Thn is a subgroup cb H AB G H thus a + ab AB = b again sinc H Sinc multiplication of matrics is an associativ law of composition, G is a group Easy computation rvals that if w lt X = 0 and 0 Y = 0 4 thn X, Y, G with X =, Y = 3 = and YX X Y Find Sylow -subgroups in th following cass: a) D 0 a) Th Sylow subgroup of D 0 ar 6 {, x, y, x y}, {, x, xy, x y}, {, x, x y, x y} 3 4 9 {, x, xy, xy} and {, x, xy, xy } b) T
MAT 444 Barclo Homwork 6 Pag 6 b) Sinc T = = 3, th Sylow -subgroups hav ordr = 4 First not that for vry dg of th thahdron thr is xactly on dg not incidnt to it, calld such pair of dgs opposit dgs Considr th rotation of 80 about th lin passing through th cntr of opposit dgs, ach of ths rotations has ordr, thr ar 3 of thm and togthr with th idntify thy form a subgroup of T of ordr 4 You can show that this subgroup is normal, thus th only Sylow subgroup d) I d) I = 60 = 3 Not that bsids th idntity thr ar lmnts whos ordr divids 4; thy ar th non-trivial stabilizr of th dgs Thy all hav ord and ar conjugat to on anothr Sinc thy ar th only non-trivial lmnts whos ordr divids 4, th Sylow -subgroups must contain som of thos So 3 of thm togthr with th idntity must form a group of ordr 4 On can show that thr ar in fact distinct Sylow -subgroups 3 Prov that if G has ordr n = p a whr a < p and, thn G has a propr normal subgroup Lt G = p a whr a< p and Not that th conjugat of a Sylow p-subgroup is a Sylow p-subgroup So if thr is a uniqu Sylow p-subgroup it must b normal By th 3 rd Sylow thorm np a and np (mod p) whr n p is th numbr of Sylow p-subgroups Sinc a < p and np a < p, w hav that n p = So w hav on Sylow p-subgroup of ordr w showd that G was not simpl If a >, thn th Sylow p-subgroup is a propr subgroup of G and G is not simpl 4 Prov that th only simpl groups of ordr < 60 ar groups of prim ordr p If a =, If p is a prim intgr w hav alrady shown that any group G of ordr p is simpl So all groups of ordr:, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 3, or 9 ar simpl Combining xrciss 643, 64, 643 and 6 w hav shown that all groups whos ordr is not qual to 4, 30, 36, 40, 48 and 6 ar not simpl Thus, w ar lft to show that ths groups ar not simpl
MAT 444 Barclo Homwork 6 Pag First not that if G = pm whr p is a prim that dos not divid m, thn if S and T ar distinct Sylow p-subgroups S T = {} Indd, sinc S = T = p thn nithr S and T hav non-trivial propr subgroup, thus S T S and T has to b trivial Similarly, if S and T ar Sylow p- and q-subgroups, thn S T S and T so thr would b lmnts of ordr q in S and of ordr p in T which is clarly impossibl Hnc S T = {} Cas : G = 4 has bn dalt with in xrcis 64 (c) Cas : G = 30 = 3 Lt n p dnot th numbr of Sylow p-subgroups W hav that n 6 and np mod which mans that n = or6 If n =, thn th Sylow -subgroup is normal, thus G is not simpl If n = 6, thn ach Sylow - subgroup has 4 distinct lmnts for a total of 4 lmnts, thus w ar lft with a total of 6 lmnts including th idntity to mak th Sylow -subgroup and th Sylow 3-subgroup Thus, w must in fact hav 3 Sylow -subgroups and Sylow 3- subgroup, or Sylow 3-subgroups and Sylow -subgroup In both cass this yilds that w hav a propr non-trivial normal subgroup 3 Cas 3: G = 40 = So n must divid 8 and b congrunt to modulo This implis that n = and this Sylow -subgroup is normal 3 Cas 4: G = 6 = W hav that n 8 and n (mod), so n = or8 Again, if n =, th Sylow -subgroup is normal and w ar don If not, w would hav 8( ) = 48 lmnts laving 8 lmnts for th Sylow -subgroups of ordr 8 thus, thr is only on Sylow -subgroup yilding onc mor that G is not simpl Cas : G = 36 and cas G = 48 ar tratd in th sam mannr via th following lmma Lmma: Thr is no non-ablian simpl group G of ordr G = p m whr p is a prim that dos not divid m nor ( m )! Proof: Assum that such a group G xists By th first Sylow thorm, G contains a subgroup P of ordr p, hnc of indx m W may assum that m >, for non-ablian p- groups ar nvr simpl On can show that thr xists a homomorphism ϕ : G Sm with kr( ϕ) P Sinc w assum that G is simpl it has no propr normal subgroups; hnc kr( ϕ ) = {} and φ is an injction; that is, G ϕ( G) Sm So by Lagrang s thorm p m m! and so p ( m )! contradicting th hypothsis