Problem Do any of the following determine homomorphisms from GL n (C) to GL n (C)?

Transcription

1 Homework 8 solutons. Problem Whch of the followng defne homomomorphsms from C\{0} to C\{0}? Answer. a) f 1 : z z Yes, f 1 s a homomorphsm. We have that z s the complex conjugate of z. If z 1,z 2 are two complex numbers, then (z 1 z 2 ) = z 1z 2. Ths s exactly the statement that f 1 (z 1 z 2 ) = f 1 (z 1 )f 1 (z 2 ). b) f 2 : z z 2 Ths s also a homomorphsm. Note that f z 1,z 2 C \ {0} then (z 1 z 2 ) 2 = z 2 1z 2 2, whch s exactly the statement that f 2 (z 1 z 2 ) = f 2 (z 1 )f 2 (z 2 ). c) f : z z Ths s not a homomorphsm. If z 1,z 2 C\{0} then f (z 1 z 2 ) = z 1 z 2 whle f (z 1 )f (z 2 ) = z 1 z 2 = z 1 z 2. d) f : z z Ths s a homomorphsm. If z 1,z 2 C\{0} then z 1 z 2 = z 1 z 2, whch s exactly the statement that f (z 1 z 2 ) = f (z 1 )f (z 2 ). Problem Do any of the followng determne homomorphsms from GL n (C) to GL n (C)? Proof. a) A A t Ths s not a homomorphsm because f A,B GL n (C), then (AB) t = B t A t, and matrces n GL n (C) do not commute. So f we choose any two matrces for whch AB BA, we would get that A t B t B t A t. b) A (A 1 ) t Ths s a homomorphsm because f A,B GL n (C), then ((AB) 1 ) t = (B 1 A 1 ) t = (A 1 ) t (B 1 ) t. c) A A 2 Ths s not a homomorphsm because matrces don t commute. That s, f A,B GL n (C), then (AB) 2 = ABAB. Ths s the same as A 2 B 2 ff AB = BA. So for any choce of non-commutng A,B we see that the map fals to satsfy the homomorphsm condton. d) A A Ths s a homomorphsm. To see ths, note that f a,b,c,d are four complex numbers, then (ab+cd) = a b +c d. Thus f A,B GL n (C), then (AB) = A B because the entres of AB are just sums and multples of the entres of A and B. And ths s exactly the formula that means the map s a homomorphsm. Problem Show that a functon φ : G G s a homomorphsm f and only f {(g,φ(g)) g G} s a subgroup of G G. Proof. Let G,G be two groups. Suppose some functon φ : G G s a homomorphsm, and consder the subset H = {(g,φ(g)) g G} of G G. We show that H s a subgroup by verfyng the followng propertes. Closed under multplcaton: Suppose(g, φ(g)) and(h, φ(h)) are elements of H. Then (g,φ(g))(h,φ(h)) = (gh,φ(g)φ(h)) 1

2 2 But φ s a homomorphsm, so φ(g)φ(h) = φ(gh). So n fact (g,φ(g))(h,φ(h)) = (gh,φ(gh)) Thus (g,φ(g))(h,φ(h)) H, so H s closed under multplcaton. Identty: Snce φ s a homomorphsm, φ(e) = e where e s the dentty of G and e s the dentty of G. Thus (e,e ) s n H. Snce (e,e ) s the dentty n G G, the dentty s n H. Inverses: Snceφsahomomorphsm, φ(g 1 ) = (φ(g)) 1. Thusf(g,φ(g)) H then (g 1,(φ(g)) 1 ) H. But (g 1,(φ(g)) 1 ) s the nverse of (g,φ(g)) n G G. So H has nverses. Assocatvty: Snce H s a subset of the group G G, ts elements are assocatve because elements of G G are assocatve. Therefore f φ s a homomorphsm then H s a group. Now suppose φ : G G s just some map for whch the set H = {(g,φ(g)) g G} s a subgroup of G G. We need to show that φ s actually a homomorphsm. To check ths, we need to show that for any two elements g,h G we have that φ(gh) = φ(g)φ(h). So choose two elements g,h n G. From the way that H s defned, we know that (g,φ(g)) and (h,φ(h)) are n H. Snce H s a group, we know that ther product s also n H. So the element (g,φ(g))(h,φ(h)) = (gh,φ(g)φ(h)) s n H. But the only element of H wth gh n the frst coordnate s the element (gh,φ(gh)). That s because each element of G appears as the frst entry n exactly one element of H. So we must have that (gh,φ(gh)) = (gh,φ(g)φ(h)). But that means exactly that φ(gh) = φ(g)φ(h). Therefore, f H s a subgroup of G G then φ s a homomorphsm. So we are done. Problem Let G be the subgroup of S 8 generated by (12)(5) and (78). Then G acts as a group of permutatons of the set X = {1,2,...,8}. Calculate the orbt and stablzer of every nteger n X. Proof. Note that the permutatons α = (12)(5) and β = (78) are dsjont. In fact, α permutes the numbers 1 though 5 amongst themselves, and β permutes the numbers 7 and 8. Snce nether α nor β move the number 6, the orbt of 6 s just {6}. Snce β takes 7 to 8 and 8 back to 7, the orbt of both 7 and 8 s the set {7,8}. Fnally, α s the product of two dsjont cycles. So the set {1,2,} s one orbt, and the set {,5} s another orbt. To compute stablzers, note that snce α and β are dsjont, every element of G can be wrtten as a power of α tmes a power of β. An element α n β m leaves a number fxed ff α n () = and β m () =. So every element of G leaves 6 fxed. Only powers of β leave the numbers 1 through 5 fxed, and only powers of α leave the numbers 6 and 7 fxed. So the stablzer of the numbers 1 though 5 s the subgroup generated by β, the stablzer of 6 s G and the stablzer of the numbers 6 and 7 s the subgroup generated by α. Problem The nfnte dhedral group D acts on the real lne n a natural way (see Chapter 5). Work out the orbt and the stablzer of each of the ponts 1, 1 2, 1. Answer. Recall that D s generated by the functons t : R R s.t. t(x) = x+1 and s : R R s.t. s(x) = x. Every element of D can be wrtten as s t n where = 0,1 and n Z. Recall further that t n (x) = x+n (for any n Z, so n can be

3 negatve) and st n (x) = x n. Translatons of the form t n move every number, so they cannot be n the stablzer of any number. The reflecton st n moves every number except for the x that satsfes x = st n (x). That s, st n moves every number except the x s.t. x = x n. We compute that ths x s n/2. Orbt of 1: The set of all ntegers. Snce elements of D send ntegers to ntegers, the orbt of 1 s at most the set of all ntegers. Snce t n (1) = n+1, all ntegers are n fact n the orbt of 1. Stablzer of 1: From the general dscusson at the begnnng of ths soluton, we see that the only elements that fx 1 are the dentty e and the reflecton st 2. So the stablzer of 1 s < st 2 >= {e,st 2 }. Orbt of 1/2: All numbers of the form n+1/2 for n Z. To see ths, note that t n (1/2) = 1/2 + n and st n (1/2) = 1/2 n. Both of these can be wrtten as m+1/2 for some nteger m. Stablzer of 1/2: Agan, referrng to the paragraph at the begnnng of the answer, we see that the only elements of D that fx 1/2 are the dentty e and the reflecton st 1. So the stablzer of 1/2 s the group < s t 1 >= {e,st 1 }. Orbt of 1/: All number of the form n + 1/ and n + 2/ for n Z. To see ths, note that t n (1/) = 1/+n and st n (1/) = 1/ n whch can be rewrtten as st n (1/) = n 1+2/ where n and n 1 can be any nteger. Stablzer of 1/: The only element of D that fxes 1/ s the dentty e snce 1/ cannot be wrtten as n/2 for any nteger n. So the stablzer of 1/ s {e}. Problem 17.. Identfy S wth the rotatonal symmetry group of a cube as n Chapter 8, and consder the acton of A on the set of vertces of the cube. Fnd the orbt and the stablzer of each vertex. Proof. Let G be the rotatonal symmetry group of a cube. We dentfy G wth S as follows. Gven vertex v on a cube C, let v be the vertex of C farthest away from v. Then the lne segment wth endponts v and v s called a prncpal dagonal. C has four such prncpal dagonals. Snce the rotatons of a cube preserve dstance, a rotatons r wll send the farthest vertex from v to the farthest vertex from r(v). So rotatons permute the prncpal dagonals. Numberng the dagonals 1 through, we get that each permutaton of the dagonals corresponds to a permutaton of the numbers 1 through. So we get the somorphsm φ : G S that maps a rotaton to the correspondng permutaton of the numbers 1 through. Let d be the th dagonal, and number ts endponts and, so that the vertces of the cube are ether numbered or for some = 1,2,,. Now a rotaton of a cube can have one of three types of axes of symmetry. There are axes that go through the centers of opposte faces of C, axes that go through the centers of opposte edges, and axes that go through the centers of opposte vertces. (Here, opposte means farthest away.)

4 Rotatons through axes gong through opposte faces correspond to powers of -cycles. A -cycle s not n A, but the square of a -cycle s a permutaton of type (2,2), whch s n A. Rotatons through axes gong through opposte edges correspond to 2-cycles. So these rotatons do not get mapped to A. The rotatons whch axes that go through opposte vertces correspond to -cycles, whch are n A. Sowejustneedtoconsdertheactonoftwotypesofrotatons. Type1: rotatons of 180 o about axes gong through pars of faces, and Type 2: rotatons of 120 o and 20 o about axes gong through vertces. There are axes that go through opposte faces. See Fgure 1. Rotatons of 180 o about these axes correspond to the elements (12)(),(1)(2) and (1)(2). We see that (12)() (whch s rotaton about the axes on the left) corresponds to the permutaton (12 )(1 2)( )( ) of the vertces. Next, (1)(2) corresponds to the permutaton (1)(2)(1 )(2 ). Fnally, (1)(2) corresponds to (1 )(1 )(2 )(2 ). ' ' ' ' ' ' Fgure 1. Three axes of type 1. Rotatons by 180 o correspond to elements of A. Next, there are axes of type 2, and each one corresponds to a dstnct -cycle n A. In the permutaton group of the vertces, they correspond to products of 2 dsjont -cycles. For example, the -cycle (12) n A corresponds to (12 )(1 2). The -cycle (12) s just the square of (12), so the correspondng element n the permutaton group of the vertces s just the square of (12 )(1 2). Thus we only need to gve the elements n the permutaton group of the vertces that correspond to the -cycles (1), (2) and (12). The remanng -cycles are just the squares of these, so the correspondng permutatons of the vertces wll just be squares as well. The permutaton correspondng to (1) s (1 )(1 ), the permutaton correspondng to (2) s (2 )(2 ) and the permutaton correspondng to (12) s (1 2)(12 ). See Fgure 2 below for an llustraton of the axes of rotaton of type 2. Problem 17.. Gven the acton of G on a set, show that every pont of some orbt has the same stablzer f and only f ths stablzer s a normal subgroup of G. Proof. Suppose G acts on a set X. Let O = {x 1,...,x n } be an orbt of the acton. Suppose the stablzer of each of the x s the same group H. We need to show that H s normal. So let h be an element of H, and let g be any other element of G. Consder the acton of ghg 1 on an element x. We know that g 1 x s some other element

5 5 ' ' ' ' ' ' ' ' Fgure 2. Four axes of type 2. Rotatons by 120 o,20 o correspond to elements of A. x j n O. Then h s n the stablzer of every element of O, so h x j = x j. Fnally, g 1 x j = x by the propertes of an acton. So ghg 1 x = x. Therefore f h s n the stablzer of x, so s ghg 1. Snce the stablzer of x s H, we have that ghg 1 = H, so H s normal. Now suppose O = {x 1,...,x n } s an orbt of the acton of G on X. Let S(x ) be the stablzer of x. We need to show that f S(x ) s normal for all then all of the S(x ) are the same. Let g be an element of G for whch g x = x 1. There s such an element of g snce x 1,x are n the same orbt. Suppose h S(x 1 ). Then g hg 1 s n S(x 1 ) snce S(x 1 ) s normal. So g hg 1 x 1 = x 1 But g 1 x 1 = x by the propertes of an acton. So that formula really means that g h x = x 1. Multplyng both sdes by g 1 that we get that h x = g 1 x. But ths just means h x = x So h, whch was n S(x 1 ), s also an element of S(x ). Thus S(x 1 ) S(x ). Snce we only used that S(x 1 ) s normal, and snce S(x ) s also normal, the same reasonng also gves us that S(x ) S(x 1 ). So all of the S(x ) are equal. Problem If G acts on X and H acts on Y prove that G H acts on X Y va (g,h)(x,y) = (g(x),h(y)) Check that the orbt of (x,y) s G(x) H(y) and that ts stablzer s G x H y. We shall call ths acton the product acton of G H on X Y. Proof. G H acts on X Y f for any two elements (g,h),(g,h) G H and any element (x,y) X Y we have that ( (g,h)(g,h) ) (x,y) = (g,h) ((g,h) (x,y) ) and f the dentty (e,f) acts on any (x,y) X Y trvally, so (e,f) (x,y) = (x,y).

6 6 The frst condton holds because ( (g,h)(g,h) ) (x,y) = (gg,hh ) (x,y) = ((gg ) x,(hh ) y) = (g (g x),h (h y)) because G,H act on X,Y = (g,h) (g x,h y) = (g,h) ((g,h) (x,y) ) The second condton holds because the dentty n G H s (e,f) where e s the dentty n G and f s the dentty n H. So (e,f) (x,y) = (e x,f y) and ths s just (x,y) snce G,H act on X,Y. Thus we have shown that G H acts on X Y. Let G(x), H(y) be the orbts of x,y n G,H, respectvely. Then the orbt of (x,y) n G H s the set G H(x,y) = {(g,h) (x,y)}. Ths s exactly the set {(g x,h y) g G,h H}. Thus G H(x,y) = G(x) H(y). Let G x,h y be the stablzers of x,y n G,H, respectvely. Then the stablzer n G H of (x,y) s just the set G H (x,y) = {(g,h) (g,h) (x,y) = (x,y)}. Ths s exactly the set {(g,h) (g x,h y) = (x,y)}. Note that (g x,h y) = (x,y) ff gcdotx = x and h y = y. Thus the stablzer of (x,y) s just G x H y. Problem Let x be an element of a group G. Show that the elements of G whchcommutewthxformasubgroupofg. Thssubgroupscalledthecentralzer of x and wrtten C(x). Prove that the sze of the conjugacy class of x s equal to the ndex of C(x) n G. If some conjugacy class contans precsely two elements, show that G cannot be a smple group. Proof. Let C(x) denote the set of elements of G that commute wth x. We frst need to show that ths s a group. The set s closed under multplcaton because f g,h commute wth x then (gh)x = g(xh) = x(gh), so gh commutes wth x. The dentty s n C(x) because the dentty commutes wth everythng. If hx = xh, then multplyng both sdes by h 1 on the rght and on the left gves us that xh 1 = h 1 x, so h 1 C(x). Lastly, C(x) s assocatve because any subset of the group G s assocatve. Therefore C(x) s a subgroup of G. The ndex of C(x) s the number of cosets of C(x). Suppose {C(x),g 1 C(x),..., g n C(x)} s the set of dstnct cosets. Say C(x) = g 0 C(x) where g 0 s the dentty. That way, all the cosets are of the form g C(x) for some. Now suppose g,g are two elements of g C(x). Then g = g h where h C(x). So g xg 1 = (g h)x(h 1 g 1 ) = g xg 1 sncehcommuteswthx. Soallelementsofg C(x)conjugatextothesamenumber. Thus x has at most as many elements n ts conjugacy class as C(x) has cosets. We wll show that each g conjugates x to a dfferent number. Suppose not. That s, suppose g C(x) g j C(x) but g xg 1 = g j xg 1 j. Then manpulatng ths equaton gves us that g 1 j g x = xg 1 j g. But ths means exactly that g 1 j g C(x). Recall from a prevous problem set that n ths case, g C(x) = g j C(x). Ths contradcts our assumpton to the contrary, so we have that all elements of the form g xg 1 are dstnt. Therefore, C(x) has exactly as many cosets as x has elements n ts conjugacy class. That means that the sze of the conjugacy class of x s the ndex of ts centralzer.

7 7 Suppose that there s an x whose conjugacy class has exactly two elements. That means the C(x) has ndex 2. But all ndex 2 subgroups of a group are normal. So G has a proper normal subgroup (that s nether {e} nor G.) That means G cannot be smple.