Math 594. Solutions 1
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1 Math 594. Solutons 1 1. Let V and W be fnte-dmensonal vector spaces over a feld F. Let G = GL(V ) and H = GL(W ) be the assocated general lnear groups. Let X denote the vector space Hom F (V, W ) of lnear maps from V to W, but vewed only as a set (.e., we gnore that X has a natural structure of F -vector space va pontwse operatons). Recall that X has a natural left H-acton and rght G-acton through composton and nner composton respectvely (.e., h.x = h x whle x.g = x g). () If x = x g for some g G then snce g GL(V ), we have gv = V and therefore x(v ) = x (V ). Conversely, suppose that x(v ) = x (V ) = U W. We frst take a bref asde: Any surjectve map f : V U of fnte dmensonal vector spaces nduces an njectve map f : U V va f (a) = a f. Injectvty of f follows from the fact that gven a b U we can fnd u U such that a(u) b(u), and, snce f s surjectve, there exsts v V wth f(v) = u and hence f (a)(v) f (b)(v),.e. f (a) f (b). Now snce x(v ) = x (V ) = U we have two embeddngs x : U V and x : U V. Ths gves an automorphsm of V takng x (U ) to x (U ). Takng the dual of ths automorphsm gves the requred automorphsm of V, usng the dentfcatons V = V, U = U, and W = W. () We clam that two elements of X le n the same H orbt f and only f they have the same kernel n V. On drecton s clear: f x = hx and v ker(x) then 0 = hx (v). But snce h GL(W ), t has kernel 0 so that x (v) = 0,.e. ker(x) ker(x ). Symmetry then gves equalty. Conversely, suppose that ker(x) = ker(x ) = U V. Then we get two embeddngs x : V/U W x : V/U W whch nduce an automorphsm of x(v ) W that can lfted (agan by extendng a bass of x(v ) to a bass of W ) to an automorphsm h of W satsfyng x = hx. () In the case W = F we have X = Hom F (V, F ) = V. If x V s not the zero map, then there exsts v V such that x(v) = α 0 so that x(λv) = λα for all λ F,.e. the mage of V under x s all of F. It follows by part () that there are precsely two orbts of G on V, namely, the orbt consstng of only the zero map, and the orbt consstng of all nonzero maps. In other words, G acts transtvely on the set of lnes n V. But as we show n Problem 3 (), the set of lnes n V s dentfed wth the set of hyperplanes n V, and ths dentfcaton s compatble wth the acton of G, so that G acts transtvely on ths set also. 2. Let F be a feld, and defne S 1 (F ) = {(a, b) F 2 a 2 + b 2 = 1}. () In the case F = R, there s an obvous way to add ponts on the unt crcle: namely, addton of angles. Recall that S 1 (R) s parameterzed by (cos(θ), sn(θ)) and that we have the addton formulae cos(θ + φ) = cos(θ) cos(φ) sn(θ) sn(φ) sn(θ + φ) = sn(θ) cos(φ) + cos(θ) sn(φ). Ths suggests endowng S 1 (F ) wth a group structure by the rules and (a, b)(c, d) = (ac bd, ad + bc) (a, b) 1 = (a, b). It s not dffcult to check that ths makes S 1 (F ) nto an abelan group (one must check assocatvty and commutatvty) wth dentty (1, 0). () Suppose that char(f ) 2 and defne φ : S 1 (F ) F by φ (a, b) = a + b. Observe that φ ((a, b)(c, d)) = (ac bd) + (ad + bc) = (a + b)(c + d) = φ (a, b)φ (c, d) so that φ s a group homomorphsm snce clearly φ(1, 0) = 1. Now f φ(a, b) = a + b = 1 then snce (a + b)(a b) = 1 we also have a b = 1 and therefore a = 1, b = 0 snce 2 0 n F. Thus, φ s njectve. Now suppose that x F \ {0}. Let a = 1 ( x + 1 ) b = 1 ( x 1 ). 2 x 2 x 1
2 2 Then we have a + b = x and a b = 1/x and hence (a, b) S 1 (F ). Snce x F was chosen arbtrarly, φ s surjectve and hence that φ s an somorphsm. Wth φ = a b, observe that φ φ 1 (x) = x 1. Meanwhle, f char(f ) = 2, defne φ : S 1 (F ) F by φ(a, b) = b. We have φ((1 b, b)(1 d, d)) = (1 b)d + (1 d)b = b + d = φ(1 b, b) + φ(1 d, d), so that φ s a group homomorphsm as φ(1, 0) = 0. The map φ s certanly njectve as φ(a, b) = 0 f and only f (a, b) = (1, 0). To see surjectvty, observe that a 2 + b 2 = (a + b) 2 = 1 f and only f a + b = 1 so that every (a, b) S 1 (F ) satsfes a = 1 b, and t s clear that (1 b, b) S 1 (F ) maps to b F under φ. Thus, φ s an somorphsm. () In a word: No. Snce we have a feld automorphsm carryng to, the feld structure cannot tell the dfference between the two. 3. Let F = F p denote the feld wth p elements for a prme p (.e., the ntegers mod p). Let V be an n-dmensonal vector space over F, and G = GL(V ). Note that V has sze p n (as one sees by choosng a bass). Assume n > 0. () Let Y = V {0}. Then Y = p n 1. Moreover, F acts on Y by scalar multplcaton, and the orbts are precsely the non-zero parts of the lnes, whch consst of exactly p 1 elements (the nonzero scalar multples of a gven vector). Snce each orbt has sze p 1, there are (p n 1)/(p 1) orbts,.e. (p n 1)/(p 1) lnes n V. Now, for any hyperplane H V, consder the subspace U of V gven by U = {f V : f(h) = 0}. Ths s clearly a one dmensonal space snce any f U s a map from V to V wth kernel H so that we have the (noncanoncal) dentfcaton U V/H F. But a one dmensonal subspace of V s just a lne n V so that hyperplanes n V correspond to lnes n V. Snce V has dmenson n, there are (p n 1)/(p 1) hyperplanes n V by our consderatons above. () Snce G acts transtvely on the set X of hyperplanes, for any hyperplane x 0 X, the orbt of x 0 under G s all of X. Snce G = Orb G (x 0 ) Stab G (x 0 ), we fnd that G = X Stab G (x 0 ). Now we can extend any bass of x 0 to a bass for V and therefore, snce any h GL(x 0 ) must take a bass to a bass, we can extend h to some g GL(V ) wth g x0 = h. It follows that the map Stab G (x 0 ) GL(x 0 ) s surjectve, and obvously has kernel Fx G (x 0 ). Thus we have Stab G (x 0 ) = GL(x 0 ) Fx G (x 0 ). () From (), we have X = (p n 1)/(p 1). Now any f Fx G (x 0 ) must act as the dentty on x 0. We let v 0 be any vector complementary to x 0. Then f can take v 0 to any vector not n x 0 snce the only requrement on f s that t take a bass of V to a bass of V. Snce x 0 = p n 1 and V = p n, there are exactly p n p n 1 choces for the mage of v 0, and each such choce specfes a unque f Fx G (x 0 ). Thus, Fx G (x 0 ) = p n p n 1. Thus, G = pn 1 p 1 (pn p n 1 ) GL(x 0 ) = p n 1 (p n 1) GL(x 0 ). Snce for any one dmensonal space W we have GL(W ) = (p 1), nducton gves G = p j 1 (p j 1) = p 2j (n+1) (p n p n j ) = p 2 n(n+1) 2 n(n+1) (p n p n j ) = (p n p n j ). (v) There are p n 1 ways to specfy the frst column of any nvertble n n matrx wth entres n F p snce the only requrement s that t not be the zero vector. Once ths s chosen, we can specfy any vector for the second column that s not n the span of the frst. Snce the span of the frst vector s just ts scalar multples, there are p n p choces for the second column. The thrd column can be any vector not n the span of the frst two columns; there are p n p 2 such vectors, and so on. Ths gves G = (p n p n j ), exactly as n (). 4. Consder the acton of G = GL 2 (F 3 ) on the set X of all 4 lnes n V = F 2 3. Let ρ : G Aut(X) be the acton map for the natural left acton of G on X.
3 3 () The four lnes n V are the scalar multples of the followng four vectors: { ( ( ( } x 1 =, x 0) 2 =, x 1) 3 =, x 1) 4 =. 1 We choose the orderng gven above. My favorte sx elements of G are a = b = d = e = c = f =. It s clear that none of these 6 elements are scalar multples of eachother. Computng the cycle decomposton of ρ(g) for the above sx g s routne but tedous. We llustrate wth a: One checks that over F 3 we have ax 1 = x 1 ax 2 = x 3 ax 3 = x 4 ax 4 = x 2, from whch t follows that ρ(a) = (234). We summarze these calculatons n the table below: g a b c d e f ρ(g) (234) (12)(34) (123) (134) (14)(23) (12) () Suppose that ρ(g) = 1. Then g fxes every lne n V and thus, for each v V we have gv = λ v v for some scalar λ v possbly dependng on v. However, let w, v V be two lnearly ndependent vectors (f dm(v ) = 1, the asserton s trval). Then snce g GL(V ) we have g(v + w) = λ v+w (v + w) = gv + gw = λ v v + λ w w, and the lnear ndependence of v, w forces λ v+w = λ v = λ w. It follows that λ v = λ s ndependent of v and gv = λv for all v V,.e. g s scalar multplcaton. Snce F 3 = ±1, t follows that the kernel of ρ s ±1 so that ρ(g) = ρ(g ) f and only f g = ±g. () By Problem 3 () or (v) we know that GL 2 (F 3 ) = 48. Snce S 4 = 24 and ker(ρ) = 2, we see that ρ s surjectve. (v) Extra credt: We choose coordnates n V and suppose that we have three (dstnct) lnes gven by the span of the three (nonproportonal) vectors a1 a2 a3 v 1 =, v b 2 =, v 1 b 3 =. 2 b 3 We clam that there s a 2 2 nvertble matrx x y α = z w such that αe 1 = λ 1 v 1, αe 2 = λ 2 v 2, α(e 1 + e 2 ) = λ 3 v 3 for scalars λ and e the standard bass. Indeed, t s enough to fnd λ satsfyng λ 1 a 1 + λ 2 a 2 = λ 3 a 3 λ 1 b 1 + λ 2 b 2 = λ 3 b 3 as you wll check. Ths s clearly possble snce we have two equatons n three unknowns. Moreover, snce the three lnes we started out wth are dstnct, we obtan an nvertble matrx. Fnally, snce we can take e 1, e 2, e 3 to any three nonproportonal vectors, GL(V ) acts trply transtvely on the lnes n V. 5. Let G be a group. () Let n be the least postve nteger such that g n = 1, and suppose that g m = 1 for some m Z. Snce Z s Eucldean, we may wrte m = nq + r for some q Z and 0 r < n. We then fnd that 1 = g m = g nq+r = (g n ) q g r = g r. If r > 0, ths contradcts our choce of n. It follows that r = 0 and n m. Now f d n and g has order n then g d has order n/d. To see ths, observe that (g d ) (n/d) = g n = 1 and f there exsts 0 < r < n/d wth (g d ) r = 1 then rd < n, contradctng the fact that n s the order of g.
4 4 () There are lots of examples. Here s one: Consder the group SL 2 (Z) of 2 2 nteger matrces of determnant 1. One easly checks that 0 1 A = and B = 1 1 are elements of G satsfyng A 2 = B 3 = 1. However, AB = clearly has nfnte order. () Suppose that G s cyclc of order n and let g 0 be a generator of G. Snce every element of G s of the form g j 0 for some j, we have, for any g = g 0, h = g j 0 G, gh = g 0g j 0 = g+j 0 = g j 0 g 0 = hg so that G s abelan. Moreover, defne φ : Z/n G by φ( mod n) = g0. Observe that ths s well defned snce t does not depend on the choce of representatve : ndeed, f j mod n then for some nteger m we have j = + mn so that g j 0 = g+mn 0 = g0(g 0 n ) m = g0. Ths s obvously a homomorphsm as g +j 0 = g0g j 0. Moreover, t s njectve snce f g0 = 1 then by part () we have n, that s, 0 mod n. Snce Z/n and G both have sze n, φ s an somorphsm. It follows that up to non-canoncal somorphsm (observe that we had to choose a generator) there s one cyclc group of order n for each postve nteger n, e.g. Z/n. (v) Let G be cyclc of order n. Let φ : G G be the map φ (g) = g. We have φ (gh) = (gh) = g h = φ (g)φ (h) snce G s abelan. Snce φ (1) = 1, we see that φ s a homomorphsm. Now suppose that φ (g) = 1 for some g G. Then f d s the order of g we see that d. Thus, f (, n) = 1 we must have d = 1 snce d n, whence g = 1. Conversely, suppose that (, n) = m > 1. Part () shows that there s a g 1 G of order n/m, so that φ (g 1 ) = 1. Therefore, φ s njectve (and hence an somorphsm by a countng argument) f and only f (Z/n). Fnally, suppose that ϕ End group (G) and let g 0 G be a generator. It s clear that ϕ s completely determned by ϕ(g 0 ). Moreover, we must have ϕ(g 0 ) = g0 for some, whence ϕ(g) = g for all g G. It follows that every endomorphsm of G s of the form φ for some. But we have seen that φ s an automorphsm f and only f (Z/n). We thus defne a map (Z/n) Aut group (G) by mod n φ. Ths s bjectve and well defned as we have noted, and s a homomorphsm snce φ φ j (g) = g j = φ j (g) for all g G. We thus have an somorphsm of groups Aut group (G) (Z/n). Observe that ths somorphsm does not depend on any choce of generator of G. 6. Fx n > 1. For σ S n and a par {, j} of dstnct ntegers between 1 and n (nclusve), note that (σ() σ(j))/( j) = (σ(j) σ())/(j ), so ths rato does not depend on the orderng among and j. Defne ε n (σ) σ() σ(j), j where the product s taken over unordered pars of dstnct ntegers between 1 and n. () We have the followng table: In general, we have {,j} 1 (12) (13) (23) (123) (132) ε n (σ) 2 {,j} σ() σ(j) σ(j) σ() j j σ() σ(j) j,j j = 1,
5 5 snce any σ S n s a bjecton of the set {1, 2,..., n} so that for every j n {1, 2,..., n} and any σ S n, there s a unque par k l {1, 2,..., n} wth σ() = k and σ(j) = l. It follows that ε n (σ) = ±1. () We have ε n (στ) σ(τ()) σ(τ(j)) j {,j} {,j} {,j} {,j} {,j} σ() σ(j) τ 1 () τ 1 (j) σ() σ(j) j σ() σ(j) j σ() σ(j) j = ε n (σ)ε n (τ), j τ 1 () τ 1 (j) {l,k} {l,k} l k τ 1 (l) τ 1 (k) τ(l) τ(k) l k snce, agan, any σ S n s a bjecton of the set {1, 2,..., n}. Clearly 1, (12) = σ S n for all n > 1 and ε n (1) = 1 whle σ(1) σ(2) σ() σ(j) σ(1) σ(j) σ(2) σ(j) ε n (σ) = 1 2 j 1 j 2 j = = 1, j 1,2 {,j},j {1,2} 2 j 1 j 1 j 2 j snce σ = (12) fxes every j 1, 2. Hence, ε n : S n {±1} s a surjectve group homomorphsm. j 1,2 7. Let A be an abelan group, a, a A elements wth respectve fnte orders n and n. () Set n = n/p e and let a = a n shows that we must have r p e. Then certanly ape so that r = p b the p are dstnct, t follows that (p e, n ) = 1 so that p e pb order p e. An dentcal argument works for a and n. = 1. Let a have order r. Then problem 5 () = 1 so that n n p b. Snce whence e b. Hence, b = e and a has for some b e. But then a npb () Suppose that aa has order r. Then 1 = (aa ) rn = a rn (snce A s abelan) so that n rn. Snce (n, n ) = 1 we must have n r. Smlarly, n r. Agan, snce (n, n ) = 1 ths mples that nn r. But (aa ) nn = 1, so that aa has order nn. Now suppose that (n, n ) = d. Wrte n p u and n p v where the set of prmes n both products s the same but u, v are possbly 0. By part (), there exsts x of order p max{u,v} for each. Snce the p are dstnct, f follows from the above and nducton that x has order p max{u,v } = lcm(n, n ). () Let F be a commutatve feld and let x F have maxmal order m. If y F has order n then n m. Otherwse, by part () there exsts an element of order lcm(m, n) > m, contradctng the maxmalty of m. It follows that every y F s a root of X m 1 = 0. Snce ths has at most m roots, we must have F m. On the other hand, x F and by the choce of m, {1, x, x 2,..., x m 1 } are all dstnct. It follows that F = m and hence an element of maxmal order s a generator for F and F s cyclc. The generators for F 17 and F 31 are lsted n the followng table: F F
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