Solutions for Tutorial 1

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1 Toc 1: Sem-drect roducts Solutons for Tutoral 1 1. Show that the tetrahedral grou s somorhc to the sem-drect roduct of the Klen four grou and a cyclc grou of order three: T = K 4 (Z/3Z). 2. Show further that the tetrahedral grou s somorhc to the alternatng grou on four elements: T = A 4. Proof of 1 and 2. Let V be the set of vertces of the regular tetrahedron. Labelng the vertces, we obtan a bjecton V = {1, 2, 3, 4}. The acton of the tetrahedral grou on V s encoded n an njectve ma T Bj(V, V ) = S 4. Indeed, any symmetry of the tetrahedron s unquely determned by ts effect on V, hence njectvty. To determne the number of elements of T, we aly the orbt stablzer theorem to the frst vertex. Its orbt s all of V, whle ts stablzer conssts of three rotatons. Hence the order of T equals 3 4 = 12. The elements of order two of T are exactly the rotatons through the mdonts of ooste edges. There are three of them, corresondng to the ermutatons (12)(34), (13)(24) and (14)(23). Together wth the dentty, these form a subgrou N of A 4 that s somorhc to the Klen four grou. All the other elements of T are rotatons of order 3 around the axs fxng some vertex of T. They are organzed nto three cosets of N as follows: N and {(123), (134), (243), (142)} and {(132), (143), (234), (124)}. Together, these form the alternatng grou A 4. N s normal, snce the cycle decomoston s nvarant under conjugaton of ermutatons. An examle of a system of reresentatves for N s R = {e, (123), (132)}. Ths choce of R s n fact a subgrou of A 4, whch s cyclc of order 3. Ths roves that A 4 s somorhc to a sem-drect roduct as clamed n 1. For the sake of comleteness, we wll also dentfy the acton of Z/3Z = (123) on K 4 = {e, (12)(34), (13)(24), (14)(23)}. It s enough to dentfy the acton of the generator. Ths needs to fx the dentty element e, and t ermutes the other three elements as follows: (12)(34) (14)(23) (13)(24) (12)(34). 1

2 3. Show that the octahedral grou s somorhc to the sem-drect roduct of the Klen four grou and the symmetrc grou on three elements: O = K 4 S Show further that the octahedral grou s somorhc to the symmetrc grou on four elements: O = S 4. Proof of 3. and 4. The md-onts of the faces of the regular octahedron form the vertces of a cube and vce versa, so that O s somorhc to the symmetry grou of the cube. Let us assume that we have labeled the faces of the cube as s customary for dce,.e., wth ooste faces addng u to seven. The orbt of the face has sx elements, ts stablzer conssts of four rotatons, so the order of O equals 24. Ths tme the ma O S 4 s gven by the acton of O on the four long dagonals of the cube. Lets number them as follows: dagonal 1 connects the vertex ( ) wth ( ), dagonal 2 connects ( ) and ( ), dagonal 3 connects ( ) and ( ) and dagonal 4 connects ( ) and ( ). Then the rotatons fxng the faces and form the subgrou (1234), the 180 rotaton through the mdonts of the edges ( ) and ( ) s the transoston (13), the rotatons around the vertex ( ) form the subgrou (234), and so on. Insde S 4, the subgrou N s stll normal (for the same reason). There are now three more cosets: {(12), (34), (1324), (1423)} and and {(13), (1234), (24), (1432)} {(23), (1342), (1243), (14)}. A system of reresentatves s now gven by R = {e, (123), (132), (12), (13), (23)}, whch s somorhc to S 3, actng on the three non-trval elements of the four grou N by ermutatons, e.g., (12) swas (13)(24)) wth (14)(23), etc. Toc 2: Central extensons Assume we are gven a short exact sequence 1 C G G 1 such that (C) s contaned n the center of G. Ths s called a central extenson of G by C. 2

3 1. For each g G, fx a reresentatve g 1 (g). Show that the falure of the ma g g to be a grou homomorhsm s measured by a ma β : G G C. More recsely, show that there s a unque such β wth g h = (β(g, h)) gh. Proof. Snce s a grou homomorhsm, we have In other words, ( g h ( gh) 1 ) = ( g) ( h) ( gh) 1 = g h (gh) 1 = 1. g h ( gh) 1 ker() = m(). Snce s njectve, there s a unque element of C mang to g h ( gh) 1 under. Call that element β(g, h). 2. Show that β satsfes the 2-cocycle condton β(g, h)β(gh, k) = β(g, hk)β(h, k). Proof. We wll suress from the notaton and treat β(g, h) as an element of the central subgrou. Recallng that central elements commute wth all elements, we obtan β(g, h)β(gh, k) (gh)k = β(g, h) gh k = ( g h) k = g ( h k) = g β(h, k) hk = β(h, k) g hk = β(h, k)β(g, hk) g(hk). Dvdng both sdes from the rght by ghk, we obtan the desred dentty. 3. Gven a dfferent choce of reresentatves wth assocated 2-cocycle β, show that there exsts a ma γ : G C wth γ(g)γ(h)β(g, h) = γ(gh)β (g, h). 3

4 Proof start. Denote the second choce of reresentatves by g g. Throughout the argument, we suress from the notaton and vew C as a subgrou of G. Let γ(g) := g g 1. Ths s n C, because (γ(g)) = g g 1 = 1. We have β(g, h) γ(gh) β (g, h) 1 = β(g, h) ( gh) gh 1 = g γ(h) g 1 = γ(g) γ(h). Here we have used the fact that elements of C are central n G. β (g, h) 1 = g h 1 h g 1 4. Conversely, gven a grou G, and abelan grou C and β as above, construct a central extenson G β of G by C. Gven β and γ, construct an somorhsm from G β to G β. Proof. The case that s normally consdered s the scenaro where we have the addtonal denttes β(1, g) = 1 = β(g, 1) for all g G. We wll lmt ourselves to ths secal case. As a set G β := C G. The multlcaton s defned as (c, g) (d, h) := (cdβ(g, h), gh). The cocycle condton guarantees that ths s assocatve, our addtonal denttes ensure that (1, 1) s a neutral element. The ncluson sends c to (c, 1), the quotent ma sends (c, g) to g. These are grou homomorhsms. The somorhsm assocated to γ sends (c, g) to (cγ(g) 1, g). 5. Work through the examles of (a) the quaternon grou (b) the bnary tetrahedral grou (c) the bnary octahedral grou 1 {±1} Q K {±1} 2T T 1 1 {±1} 2O O 1 4

5 (d) the bnary cosahedral grou 1 {±1} 2I I 1. Proof. The central subgrou of Q s {±1}, the multlcaton s determned by the table j k 1 k j j k 1 k j 1 (frst entry n row n tmes frst entry n row m equals entry nm). Choose the set of reresentatves {1,, j, k} for, say a = {, } and b := {j, j} and ab := {k, k}. Then β s reresented by the table β 1 a b ab a b ab You can look u the elements for the other grous on Wkeda, choose reresentatves and wrte out the multlcaton tables, f you wsh. Ths s lengthy (a 60 tmes 60 table for 2I), but elementary. More on sem-drect roducts: Fnd somorhsms and 2T = Q (Z/3Z) O = T (Z/2Z). Fnally, show that there s a short exact sequence 1 Q 2O S 3 1 but no way to wrte the bnary octahedral grou as a sem-drect roduct of Q and S 3. Proof. The frst statement s an alcaton of the thrd somorhsm theorem: Consder the sequence of nclusons {±1} < Q < 2T. The quotent Q/{m1} s somorhc to the Klen four grou, the quotent 2T/{±1} s the tetrahedral grou T, and the ncluson s our old frend N from the frst set of questons. Snce N s normal n T, t follows that Q s normal n 2T. Moreover, we have an somorhsm 2T/Q = T/N = Z/3Z. 5

6 Consder a rotaton R of order 3 n T, and choose a reresentatve R of 1 (R) n 2T. Then ( R 3 ) = 1, so R has ether order 3 or order 6. In ether case, R2 has order 3 and mas to R 2 under. Hence the reresentatves for the cosets of Q nsde 2T can be chosen as the elements of the cyclc subgrou R 2 of order 3 n 2T. If you look at the Wkeda age on the bnary tetrahedral grou, you fnd a second descrton of the sem-drect roduct structure. Now the second clam s the we can fnd a slttng for the short exact sequence sgn 1 A 4 S 4 {±1} 1. Note that any subgrou of ndex two s automatcally normal. Such a slttng s gven by any choce of odd ermutaton of order 2, for nstance by (12). For the sake of comletenes, we note that the conjugaton acton of (12) on N s as descrbed above. The conjugaton acton on the remanng eght elements s (123) (132) (124) (142) (134) (234) (143) (243). Fnally, the argument that Q s normal nsde the bnary octahedral grou wth quotent somorhc to S 3 s another alcaton of the thrd somorhsm theorem, very smlar to the tetrahedral case. The only quaternon unt of order 2 s 1. In artcular, there s no subgrou of 2O that s somorhc to S 3. 6