ON GENERAL GROUP EXTENSION
by
John D•. Dixon, M.A.
A thesis submitted to the Faculty of Graduate Studies and Research in partia1_ fulfilment of the requirements for the degree of Doctor of Philosophy.
Department of Mathematics, McGill University,
Montreal. April 196~. 2
TABLE OF CONTEN~S
Page Introduction 3 1. Mixed Groups 6 2. Extension Problem 12 3. Cardinal Representations of Mp(l) 21 4. Extension Functions 27 5. Particular Extensions 35 6. Structure of Normal Extensions 44 7. Structure of General Extensions 55 Notes 60 Bibliography 63 INTRODUCTION
The problem of normal extensions in groups (see Kurosh [1] Chapt er XII) has be en studied in consider able detail and has reached the point of a reasonable solution. The corresponding probi.em of general exten sions in groups, although studied quite extensively, has not yet reached that point. The aim of the pres ent thesis is to generalise some of the resulte ob tained in the c.ase of normal extensions to the general case. It was proved by Baer t11 that the cosets of a group H modulo an arbitrary subgroup G can be given a structure called a mixed group. In partic ular when G is normal in H the mixed group is a group. Taking this generalisation of a quotient group as a mixed group as a starting point, this thesis de velops a generalisation of normal extension theory.
(cf. the alternative generalisation of Szep [~ ). Whilst no generalisation of the cohomology theory in normal extensions is considered, the structure of the different extensions of a group in the general case is examined and in particular criteria for auch an exten sion to be possible are found. The NOTES at the end of the thesis give detailed references and specify the parts of the thesis claimed to be new resulta. The author would like to express his considerable gratitude to Professor H. Schwerdtfeger who eupervised this work., Prof. Schwerdtfeger suggested the original problem of extending groups by mixed groups and he has continued to give generous help and patient encourage ment in the execution of the thesis.
Notation The following notation is used consistently: 1) (x : * 1 denotes the set of elements x which satisfy the given condition * • If A, B are two sets then AB = {ab : a E A, b E B} is defined when the products ab (a é A, b é B) are defined., In particular, if U is a subset of a group G then GU = G and u2 = U implies that U is a group when U is finite.
2) For a group H G ~ H means G is a eubgroup,
G 4 H means that G is a normal subgroup of H • The unit group is written (1) • Z(H) denotes the centre of H and N(G;H) denotes the normaliser of G in H • The order of G and the index of G in H are denoted by (G:l) and (H:G) respectively. More gen erally (M:l) denotes the order of a set M • A x B is the direct product of the groups A and B • Sm will be used to denote the symmetric group on m let ters. We may suppose Sm-l ~ Sm for m = 2, 3, ••
3) If ~ is an arbitrary mapping of a set A into a set B then a~ (a ~ A) denotes the image of a in B • If C ç A then ~ \ C denotes the mapping
~ restricted to the elements of C • If ~ is a homomorphism between the groups H and H then ker ~ denotes the iernel of the homomorphism. (Bee Kurosh l~ _s 10). The sign # denotes the end of a proof. 6
1. MIXED GROUPS
Following Baer [~ we define a mixed group ("Mischgruppe") as follows.
Definition 1.1 A mixed group is a set M for which for certain ordered pairs of elements x,y G M a prod uct xy e M is defined such that (i) there is a subset U of M which under the given product is a group called the nucleus ("Kern") of M . ' (ii).* x y is defined if and only if x e U ; in particul ar for the unit element 1 é u ly = y ; (iii) products are associative when defined, i.e. x(yz) = (xy)z when x,y E U •
*More exactly we are defining a right mixed group. A left mixed group could be defined similarly by alter ing (ii) and (iii) auch that xy is defined if and only if y é U • There is an obvious one-one correspondance between right and left mixed groups and in what follows we consider exclus ively the right mixed groups defined above.
In particular every group is a mixed group whose nucleus is the whole set. On the other hand any set can be made a mixed group by introducing a left unit 1 element and defining the product only for left mult iplication by this unit. The nucleus of this mixed group is the unit group.
Theorem 1.1 A mixed group M with nucleus U may be written as a union of disjoint cosets Ux (x E M). We write
(1.1) M = U + Ua + Ub • ••• where the sets U = U.l, Ua, Ub, •• are disjoint. Proof Each element x E M lies in one coset, viz. Ux • It is therefore sufficient to prove that any two cosets Ux, Uy are either disjoint or Ux = Uy •
Suppose Ux ~ Uy ~ ~ then there is a z ~ Ux A Uy and z = ux = vy for some u,v e.. U • But then Uz = U(ux) =(Uu)x = Ux becauee of the group property of U. Similarly Uz = Uy and so the theorem is proved. #
Definition 1.2 The (finite or infinite) cardinal number of disjoint cosets (modulo U ) in M given in the decomposition (1.1) is called the~ of M •
If M is finite with rank r we note (M:l) = r(U:l) • In particular the order of the nucleus div ides the order of the mixed group.
Definition 1.3 Two mixed groups M1 and M2 are isomorphic (written M1 ~ M2 ) if there is a mapping 8
-c ( called an isomorphism) auch that
( i) -c is one-one from the who le of M1 onto M2 (ii) the product (a~)(b~) is defined in M2 if and only if ab is defined in M1 and then (a~)(b~) = (ab) -c •
Definition 1.4 A set R containing exactly one elem- ent of M from each coset of M (mod U ) is called a set of representatives of M •
Theorem 1.2 Two mixed groups Mi with nuclei u1 and ranke ri (i = 1,2) are isomorphic if and only if the groups ul and u2 are (group) isomorphic and rl = r2 •
Pro of A. Let ""Cl be an isomorphism "t"1 : ul onto u2 and let rl = r2 • Let Rl and R2 be sets of repres- entatives for Ml and M2 respectively. Be cause rl = r 2 the re is a nne-one mapping T~ : R1 onto R2 • Each element x1 ~ Mi is in just one coset and so can be written uniquely xi = uiri (ui ~ Ui' ri 6 Ri)
1,2) • -c )-c (i = We define by (u1r 1 = (u1 L:,)(r1 -r.J (u1-c, 6 u2, r 1 ""C",_ ~ R2) • -c satisfi&S (i) of Definit ion 1. 3 a ince -c, and ""'C~ are each one-one. On the other band (x~)(y~) is defined in M2 if and only if X""t:. E:. U and hence X é Ul, X"C X 1: • Wri ting 2 = 1 y= ur (u é u1 , r é R1 ) we then have (x~)(y~) = (x -c, )( u ~. )( r ,;..) = (( xu) -c, )( r ~) ={.( xu) r 1-r: = {xy 3- -c • So (ii) of Definition 1.3 is satisfied and ~ is the required isomorphism between the mixed groups M1 and )(2 • B. Conversely, if we have an isomorphism
~ : M1 onto M2 then ~ju1 : u1 onto u2 is an isomorphism (of groups) between the nuclei of the mixed groups. ~ also gives a one-one mapping between the sets of cosets of the respective mixed groups and bence r 1 = r 2 • #
From the above theorem a mixed group is given uniquely up to an isomorphism by prescribing its nucleus
U and rank r • We shall denote this mixed group by
Mr(U) • The next theorem shows the connexion between the concept of mixed group and the problem of group exten- sion.
Theorem 1.3 Let H be a group with subgroup G 'H • We consider the set M of all cosets of H (mod G ) and define a product between cert~in ordered pairs of cosets by Gx oGy = Gz (x, y, z ~ H) if and only if the set product GxGy is a single coset Gz • M is a mixed group with nucleus U = {Gx : x é N(G;H)} where N(Q;H) is the normaliser of G in H • Proof U is a group with unit element G since for all x,y ~ N(G;H) GxGy = GGxy = Gxy and xy ~ N(G;H), 10 while Gx has inverse Gx -1 • Associativity in U follows from associativity in H • In fact U = N(G;H)/G. It remains to prove that (ii) of Defin ition 1.1 holds. Consider the set GxGy • We wish to show that
GxGy = Gz for some z ~ H if and only if x ~ N{G;H) •
If x é N{G;H) then GxGy = Gxy • On the other hand Gxy Ç GxGy so if GxGy = Gz for some z then Gxy
= GxGy • Renee Gx 2 lxG = xG and so x ~ N{G;H) • #
We shall denote the mixed group M defined in Theorem 1.3 as H/G • The nucleus of the mixed group is N{ G;H) / G and we note that for G Clearly s3;s2 ~ M3{1) • In general we have Sn/Sn-l ~ Mn{l) • Another example, in which, however, the mixed group has infinite rank, is given as follows. Define the group H = { {x, y) .• x,y real, x ~ o} with the product {x,y)(x•,y•) = {xx•, yx'• y') • It can be seen that the product is associative since {(x,y)(x•,y• )} (x",y") = (xx•, yx'+Y'Hx",y") = (:xx'x", (yx'+Y' )x" + y") whilst (x,y){(x',y')(x",y")} = (x,y)(x'x", y'x"+Y") = (xx'x", yx'x" + y'x" + y") • Furthermore, H has unit element (1,0) and the inverse -1 -1 of (x,y) is ( x , -yx ) • If we take as the subgroup G = {(x,O) : x real, x~ o} we conclude by Theorem 1.3 that H/G is a mixed group M with nucleus U = l(G;H) / G • But U = (1) since for (u,v) ~ N(G;H) for all x ~ 0 and this implies (x, v-vx) é G i.e. v - vx = 0 for all x and so v = 0 • Therefore H/G has unit nucleus and its rank is therefore the power of the continuum ( = (H:G)). -12 2.. EXTENSION PROBLEM In this and the suceeding sections we shall con aider exclusively finite mixed groups (i.e. mixed groups M wi th (M: 1) < oe ) • Some of the re sulte may be extended to the infintte case; but we shall not con- aider auch an extension. Theorem 1.3 showed that to any pair of groups G, H with G ~ H there is a corresponding mixed group, viz~ H/G • As first proved by Baer [1] we have the following converse: Theorem 2.1 With the single exception of the singular mixed group M2(l) with nucleus (l) and rank 2, there corresponds to every mixed group Mr(U) with nucleus U and rank r , a pair of groups G, H with G ~ H auch that H/G ~ Mr(U) • Proof Firstly there is no pair of groups G, H with G ~ H such that H/G ~ M2(1) • This follows immediately from the fact that a subgroup of index 2 ia normal while H/G ~M 2 (1) would imply (H:G) = 2 and N(G;H) = G • Secondly consider Mr(U) when either r ~ 2 or u 1 (1) • Let sn denote the symmetric group on the lattera 1, 2, • • n .• sl < s2 < • ••• (i) r 1 2 • Take H = u x s r ' G = sr-1• Since N(Sr_1; Sr) = sr-l for r > 2 and (Sr: sr_1 ) = r we have H/G ~ Mr(U) • (ii) r = 2, U ~ (1) • Take H • (U x U') + (U x u• )a ' G = u• where a2 = 1, a+ u and u• = { a -1 xa : Xt::U}~u. Th en N(G;H) = u x u• and (H: N(G;H)) = 2 so H/G ~ M2(U) • # Definition 2.1 A reEresentation of a mixed group M is any pair of groups G, H wi th G ~ H and H/G !::! M and is denoted by H/G • Theorem 2.2 SUppose ~ is a homomorphism of a group H onto a ~oup H' auch that a subgroup G of H is the complete set of all elements mapping onto a subgroup G• of H' • Then the mixed groups H/G and H'/G' are isomorphic and we say that the two representations are homomorphie. Proof From Theorem 1.3 it is sufficient to prove that (H:G) = (H':G') and N(G;H)/G ~ N(G';H')/G' • Let ker ~ = K then H/~ ~ H' and G/K ~ G' so (H:G) = (H ' : G' ) • Furthermore N( G; H) / K !:! N( G' ; H ' ) and so by the third isomorphism theorem Definition 2.2 Two representations H1/G1 and H2/G2 of a mixed group M are congruent if there existe a third representation H/G such that H1/G1 and H2/G2 are both homomorphie (in the sense of Theorem 2.2) to H/G • We wri te H1/G1 ,-..~ H2/G2 • Theorem 2.3 Congruence between representations of a mixed group M is an equivalence relation. If (M:l) = m every class of congruent representations contains a representation B/A where B is a transitive perm utation group of degree m • Proof Clearly congruence is a symmetric and reflexive relation. To prove that i~ is an equivalence relation we must show that it is transitive, i.e. if H1/G1 ~ H2/G2 and H /G ~ H /G then H /G ,-..~ H /G • Suppose that 2 2 3 3 1 1 3 3 H1/G1 and H2/G2 are both homomorphie to M1/L1 with kernels K1 and K2 respectively and that H2/G2 and H /G are homomorphie to M /L with kernels K2 and 3 3 3 3 K3 respectively. Then K2K2 ~ H2 , K2K2 ~ G2 so we can define H/G by H = H2/K2K2 , G = G2/K2K2 • We now show that both H /G and H /G are homomorphie 1 1 3 3 to H/G • In fact since K2K2/K2 4 H2/K2 ~ H1/K1 there is a normal subgroup Ki ~ G1 such that Ki/K1 4 H1/K1 , Ki/K1 ~ K2K2/K2 and so by the third isomorphism theorem H1/Ki ~ H2/K2Ka = H and G1/Ki ~ G2/K2K2 = G • So ~/G1 maps homomorphically onto H/G and similarly H /G maps homomorphically onto H/G so H /G ~ H /G • 3 3 1 1 3 3 This proves congruence is an equivalence relation. To prove the second part of the theorem we must show that any representation H/G of M is congruent to a representation B/A where B is a transitive permutation group of degree m = (M:l) • Since (H:G) = (M:l) = m the standard permutation representation of H over G (see Zassenhaus [11 p.40) gives the required transitive group B • Explicitly the mapping Ga. ( 2 .1 ) x ~ (Ga .~x) (x é H) ~ where H = Ga1 + Ga2 + ••• + Gam maps H homomorph- ically onto a transitive permutation group B of degree m • G map~ onto the subgroup A of B con sisting of those elements of B leaving the coset G fixed. B/A ru H/G • The kernel of the homomorphism 'V>\ l is K = (\ a-:- Ga. which is the largest subgroup of .:.-· ~ ~ G normal in H • # Corollary 2.3 For each (finit~ mixed group M there are only a finite number of classes of congruent repres entations of M • Proof Each congruence class is associated with a trans- itive subgroup of Sm (the symmetric group on m let ters) and two classes mapping onto the same transitive subgroup are identical. # Definition 2.3 of a mixed group M are equivalent if there is an ~ H H ~ • isomorphism : 1 onto 2 with : G1 onto G2 Defini ti on 2. 3 A representation H/G is called a 16 cardinal representation of a mixed group M if it is minimal in the sense that for any congruent represent- ation H1 /G 1 (H':l) ~ (H:l) • For a cardinal representation H/G the homo mmnphism defined in Theorem 2.3 onto B/A must be an isomorphism; so every cardinal representation is isomorphic to a representation B/A where B is a transitive permutation group of degree m = (M:l) = (H:G) • Conversely Theorem 2.4 Ea.ch transitive subgroup B of sm has a. subgroup A such that B/A is a cardinal representation of some mixed group M with (M: 1) = m • Pro of As is well known (see Burnside [1) § 133) every transitive subgroup B of Sm has a subgroup A of index m • Therefore B/A is a representation of some mixed group M with (M:l) = (B:A) = m • A is the subgroup of B consisting of elements of B leav- ing a given' letter fixed. Since any normal subgroup of a transitive permutation group must be transitive or the unit group the only subgroup of A normal in B is the unit group. Hence B/A cannot be homo- morphic to a representation of smaller order and so B/A is a cardinal representation. # ll Monomial groups Adopting the notation of Ore [2] we have Definition 2.5 The complete monomial group ~r(U) is the group of all one-one mappings ~ of the mixed group Mr(U) with the nucleus U and rank r onto itself with the property (2.2) (ux)-c = u(xL:) (u e. U, x E. M) • Such mappings are introduced in Baer 1 as para llel translations ("ë.hnliche Bildungen"). See also Zassenhaus [1] Chapter V in the discussion of monomial representations. Definition 2.6 A subgroup T of L r(U) is called transitive if for all x,y é M there is a 1: E. T such that X"'t = y • Because T is a group it is sufficient that for all x e.. M there is a -cc:. T auch that l"t. = x • Theorem 2.5 Every cardinal representation B/A of a mixed group M with nucleus U , rank r , has B isomorphic to some transitive subgroup T of ~r(U) • Proof We need only proTe that the mapping (2.1) of Theorem 2.3 maps the representation H/G onto a group isomorphic to a transitive subgroup of L:r(U) • -18 In fact -x = is a one-one mapping of the set of cosets of G ) onto itself, i.e. a one-one mapping of the mixed group H/G ~ M onto i.:tself. -x has the property (2.2) of Definition 2.5 sinee {(Gai)( Gaj )} x = (Gai ) { (Ga j ) i } for all x E: H • The group is transitive since (G)ai = (Gai) (for ai ~ H ) • Therefore H maps onto a transi tve sub group of L r(U) and the theorem is proved. # The Extension Problem Definition 2.7 A group H is an extension of a group G' by a mixed group M when H has a subgroup G ~ G• auch that H/G ~ M • Theorem 2.6 A group H is an extension of a group G' by a mixed group M when H has a subgroup G ~ G' if and only if H/G is homomorphie to a cardinal rep resentation B/A of M • Proof This is simply a restatement of Theorem 2.3. # The extension problem can be stated as follows: Given a group G• and a mixed group M how can we characterise the set (possibly empty) of all groups H which are extensions of G' by M ? As an example of a case where no extension existe consider the case where G' is a p-group ( p prime) and M = Mn(U) with p} (n-1) , p} (M:l) • No exten sions can exist for otherwise we should have a group H with Sylow p-group G and (H: N(G;H)) = n • But this is impossible because then H would have n con jugate Sylow p-groups and p} (n-1) contrary to the third Sylow Theorem (see Zassenhaus l~ p. 136) • We may attack the extension problem in the follow- ing manner: Let Bi/Ai (i = 1, •• k) be a complete set of inequivalent cardinal representations of the mixed group M • Then all extensions H of G' by M may be placed in classes corresponding to the congruence classes (Definition 2.2) into which H/G (G ~ G•) falls. Clearly a necessary condition that G have an exten- sion by M is that G map homomorphically onto one of the subgroups Ai of the cardinal representations. In general this condition is not sufficient as is shown by the following examp~e. Counter Exam.ple This is an example of a group G mapping homomorphically onto a group A auch that B/A is a cardinal representation of a mixed group M but no extension H of G exista auch that H/G t"V B/A • Let B = Gp { a, b, c a4 = b4 = c2 = 1, ab = ba, c -1 ac = b J = (ca ><- cb) ..- (Ca x. Cb)c 20 where ca x cb is the direct product of the cyclic groups generated by a and b • (B:l) = 32 • 2 Let A = Gp {a } • (A;l) = 2 • Then N(A;B) = Gp a,b and (N(A;B) :1) = 16. So B/A has rank 2 and nucleus N(A;E)/A • Now let G = s • Then there is exact1y one homomorphism 3 of G onto A and that has kernel K ~ Gp { u : u3= 1} • Suppose there ie an extension H of G auch - that H/G NB/A • Th en for a: ~ H with a--.aE. B --1 - --1 - u2 we have a Ka = K and so either a ua= u or • --2 -2 -2 On either case a ua = u • But a E G since it 2 -2 maps on a € A and since a:2 ~ K a and u gen- era te G and so G is an abe lian group. This is contrary to the hypothesis and so no extension H of G can exist. # We shall later give some sufficient conditions that a group have an extension conresponding to a given representation. (Notably Theorem 7.4). 21 3. CARDINAL REPRESENTATIONS FOR Mp(ll The simplest kinds of mixed groups are those whose nucleus is (1) and whose rank is a prime p • Since the only mixed groups M with (M:l) = p are either isomorphic to MP(l) or M1 (cp) where CP is the cyclic group of order p (see the note to Definition 1.2) the transitive subgroups of SP can only yield representations of one of these mixed groups. (Theorem 2.4). Since H/G is a representation of M1(cp) only if G ~ H, H/G ~CP the cardinal rep- resentations of M1 (cp) are all equivalent to C/(1) • in S Renee the cardinal representations of MP(l) p correspond to all the transitive subgroups of except for those of order p • Examples p = 2. As previously noted (Theorem 2.1) the mixed group M2(1) has no representation. p 3. The only cardinal representations of M (1) = 3 are equivalent to s3/s2 • p = 5. The cardinal representations of M5(1) are equivalent to one of the following representations: (a) ; s5;s4 22 (b) A5/A4 (where An is the alternating group on the letters 1, 2, • • • n •) ' (c) B5/B'5 where B5 = Gp {(12345), (1243)} and Gp B'5 = {(1243)} • These groups are of order 20 and 4 respectively. (d) 05/05 where 05 = Gp {(12345), ( 14 )( 23) 1 and 05 = Gp {(14 )( 23)} • These groups are of order 10 and 2 respectively. p = 7. All cardinal representations are equivalent to one of the following representations: (a) ; s7;s6 (b) A7/A6 (c) B7/B7 where B7 = Gp {(1234567), (124)(536), (1632)(45)} and B? = Gp {(124)(536), (1632)(45)} • B7 has order 24 and B is a simple group of order 7 168 (see Burnside [1] §166). ( d) c7;o7 where c7 = Gp {( 1234567), (132645)} and 07 = Gp {(132645)} • These groups have orders 42 and 6 respectively. (e) D7/D7 where D7 = Gp {(1234567)' (124 )( 365)} and D'7 - Gp {(124)(365)} • These groups have orders 21 and 3 respectively. (f) where E7/E'} E7 = Gp {(1234567), (16)(34)(25)} and E'7 - Gp {(16)(34)(25)} • These groups have orders 14 a.nd 2 respective1y. Properties of the cardinal representations of Mp(!l Let H/G be a cardinal representation of MP(l) wi th H ~ SP • From the previous remarks this me ans (i) H is a transitive subgroup of SP and (ii) (H:l) > p • 2 Since (H:G) = p and p i Theorem 3.1 If H/G is a cardinal representation of then either all the minimal normal subgroups of Mp (1) H are simple and nonabelian or H has its Sylow p-group P4H. Proof Let K 4 H , K ~ (1) • Then K ~ G because then H/K/G/K IV H/G contrary to the hypothesis that H/G is a cardinal representation. Hence KG properly contains G and so KG = H , because G is a maximal subgroup of H • In particular p 1 (K:l) since p}(G:l) and so the Sylow p-groups of H lie in K. Now suppose that K is a minimal normal subgroup of H • To prove the theorem we must show that K is simple (possibly K = P ). If K were not simple it would have a minimal normal subgroup N • Let the con- jugate groups in H be N N, N , •• N • These are 1 = 2 8 all minimal normal subgroups of K and so their prod- uct N1N2 •• Ns is a subgroup of K which is normal in H • Because K is a minimal normal subgroup K = N1N2 •• Ns • Be cause Ni' Nj are minimal normal subgroups of K the re fore (1) . Nit"\ Nj = (i ~ j ' i,j = 1, 2, •• s) and it is easily seen that there- fore K is a direct product K = N. x. N. x •• x N. of ~1 ~l ~r some subset of these conjugate groups. (cf. Burnside I~ §53). In particular (K:l) = (N:l)r so p 1 (K:l) 2 implies But p } ( K: 1 ) (because K ~ S ) Pr \ (K:l) • p and so r = 1 and K is simple. The two alternatives of the theorem arise when K contains the Sylow p-groups properly and when K equals the Sylow p-group, respect- ively. # Theorem 3.2 If H/G is a cardinal representation of MP(l) and if K is any non-trivial normal subgroup of H and if P is a Sylow p-group of H then H = N(P;H) K • Proof As noted in the proof of Theorem 3.1 the Sylow p-groups of H are in K • Hence the Sylow p-groups are the same in H and K • From the conjugacy of Sylow p-groups (Second Sylow Theorem - see Kurosh [1] §54) we have that for all x G H 1 some w ~ K • Renee xw- t N(P;H) and so x E N(P;H)w • Since this is true for all x E H H Ç N(P;H) K and the re sult follows. (cf. M. Hall [1] Theorem 4. 2. 4). # Corollary 3.2 A group H for which H/G is a card- inal representation of Mp(l) is either simple or the product of a simple normal group with the normaliser N(P;G) of a cyclic p-group P in H • Proof Since by Theorem 3.1 H = GK therefore N(P;G)K 2 N(P;H) and so H = N(P;H)K = N(P;G)K • # To complete the investigation we determine the group N(P;G) • Since we need only consider the case H ~ S it is sufficient to find N(P;S ) be cause p p-1 G S '"' H ie the subgroup of H keeping the pth = p-1 letter fixed and N(P;G) = N(P; H Asp- 1) = N(P;S p-1) 1"\ H • Furthermore we obtain a complete set of inequiv alent cardinal representations H/G even when we impose the condition that (12 •• p) E H since, as was noted at the beginning of this section, H contains at least one element of order p and every element of order p in SP is similar to (12 •• p) • Therefore we need only find N(Gp {(12 •• p)}; sp_1 ) • Theorem 3.3 N(Gp {(12 •. p)} ; Sp-l) = R where R is a regular cyclic subgroup of s p-l of order p-l • We may write R = t cr -( r = 1, 2, • • p-l) : cr : i ~ ri (mod p) ( i = 1, 2 , • • p-l) 1 . (The elements which are permuted by the permutations of Sp-l are taken to be the integers (mod p) not congruent 26 to p .) Proof Suppose x ~ N( Gp { ( 12 • • p)} ; S l) then p- x-1 (12 •• p)x = (12 •• p)r for some integer r with 1 ~ r ~ p-l • If ix denotes the image of the letter i by the mapping x then x-1 (12 •• p)x maps ix~ 1 __,. (mod p) ; while (12 •• p)r maps ix~ ix+r (mod p) • Therefore we must have (mod p) • x For i = p this gives 1 :s: r (mod p) be cause x ~ Sp-l • In general, by induction on 1 , we see that (i+l}xs (i+l)r (mod p) • Thus x= cr as def1ned above. Conversely, if x= c r it is easy to see that x 6 N(Gp {(12 •• p)J ; Sp-l) and so the theorem is proved. # Corollary 3.3 The cardinal representations of M (1) :p are all e~uivalent to representations H/G where H is a non-cyclic transitive subgroup of and is of the form H = NK where N ~ R, K~ H and K is a simple, possibly cyclic, group containing (12 •• p) while R is defined in Theorem 3.3. Thus a knowledge of the simple, transitive groups of degree p enables us to write down a complete set of cardinal representations of M {1) • p 4. EXTENSION FUNCTIONS We shall now consider the extension problem as an example of the general skew product (Redei I~ ). Let G be a group and l' a mixed group wi.th nucleus 11 • We define the skew product ( G, T' ) as the set {(a,~) : a t G, ~ ~ r} with multiplication defined by (4.1) where a, b,f(at, b,~) é G and oc ,~,tp(oc, b,~) ~ r . T.he functions f and q> wi th values in G and -r res pectively, are so far left arbitrary. The aim of the next two theoreme is to· find condit±ons which these functions must satisfy in order that the skew product ( G, l') is a group and an extension of G by l' . Note We shall take as the unit elements in G, T' and (G, l') the elements 1, 1 and (1,1) respective1y. Furthermore we shall wri te (G, 1) = {(a,l) : a ~ G} and ( 1 ' l' ) = { ( 1 ' 0(. ) : Theorem 4.1 The skew product ( G, l' ) is a group wi th unit (1,1) if and only if the fo1lowing conditions hold on the functions f and ~ for all b,c é G and ot,~,.l E. r. (4.2) f(1,b,~) = b t ~(l,b,~) = ~; (4.3) f(al,b,p)f(<\'(«.,b,\='), c, t) = 28 = f(oc, bf(f>,c,o ), cp(\?>,c,~ )) ; (4.4) q>(~(OL,b,f->), c, ~) = q>(ot, bf(~,c,i1), cp(~,c,r)) ; (4.5) For all ol. E:. r there is a se:. r auch that 9> ("! ' 1' oc) = 1 • Pro of A. Suppose S = ( G, 1' ) ie a group wi th unit (1,1) then (i) (l,l)(b,~) = (f(l,b,~), ~(l,b,~)) = (b,~) is equivalent to (4.2); (ii) Since 1_(a,o<.)(b,~)}(c,l) = ( af («. ' b ' ~ ) ' ~ (0{ ' b ' ra) )( c ' "6 ) = (af(~,b,~)f(cp(oc,b,~),c,~), ~(~(~,b,~),c,1)) and (a,ot){(b,~)(c,-g>} = (a,o<.)(bf(~,c,~), ~((3,c,~)) = ( af (oc' bf ( (3' c' ~)' q>( ~' c' i )) ' cr (o(. 'bf (13 'c '1 ) ' q>( ~' c '0 ) )) associativity in S is equivalent to (4.3) and (4.4) • (iii) Since S is a group (1,~) has an inverse (x,s) • Then (1,1) = (x,"i)(l,ol.) = (xf(J ,l,ot),q>(s,l,od) implies (4.5). B. Conversely, let the skew product S = (G,r) satisfy (4.2) - (4.5). To show that S is a group we must prove S has (1,1) as a left unit, S is associative, and each element (a,~) ~ S has a left inverse. (cf. Zassenhaus [1] p .1) • ( i) ( 1 ,1.) is a le ft unit of S from A( i) • (ii) S is associative from A(ii). (iii) (a,~) ~ S has a left inverse. To show this we note (a-1 ,l)(a,l) = (1,1) (using (4.2)) and that by (4.5) there is a ~E: l' auch that Cf(::t,l,O() = 1 • Then (a-1 ,1) = (a,1)-1 , (f(~,1,~)-1 ,~) = (1,~)-1 and so (a,~)-1 = (1,~)- 1 (a,1)-1 = (f(!,1,~)-1 f(s,a-1 ,1), ~(s,a-1 ,1)) • # Corolla.ry 4.1 If S = ( G, l') is a group wi th unit (1,1) (i.e. satisfying (4.2) -(4.5)) then (a) ( b) For all a '" G, «. E: l' there exista ~ €. r auch that Cf ('1, a,o{) = 1 ; (c) The mapping À__, cpO,,a,o<.) is a permutation of the elements of r . Froof (a) (1,~) = (1,~)(1,1) = (f(~,l,l), ~(~,1,1)) ; (b) Defining ~ by (a,~)-l = (u,~) we have (u;Ï)(a,«.) = (1,1) and so (uf(.,,a,cX.), cp(-,,a,oc:)) = (1,1) • (c) (G,1)(1,>.) ~ ( G, 1 )(1, X)(a,oc:) = (G,l)(f(À,a,~), = (G,l)(l, ~(x,a,~)) is a permutation of the right cosets (G,l)(l,À) ( >. E.. r) of: S modulo the subgroup (G,1) • Since each coset may be written uniquely in the form (G,l)(l,À) (ÀéT') this is a1so a permutation of the elements of l' . # Theorem 4. 2 If the skew product S = ( G, f1) is a group with unit (1,1) then the mixed group (G,r)j(G,l) is isomorphic to r under the natural mappi ng if and only if the following conditions hold: (4.6) ~(:r, b,l) = s (for all b E. G) if and only if "! E:; D.. • ( 4. 7) cp('!, 1, ~) =1~ when sE. ~ • (Note .6. is the nucleus of r and so sj3 is only defined when "5é6. (Definition 1.1)). Proof Since ((G,r): (G,l)) = (r:l) we have by Theorem 1.2 that (G,l')/(G,l)!::! l' if and only if the nucleus of (G,r)/(G,l), viz. N((G,l); (&,r))/(G,l) is group isomorphic to ~ • But (G,l)(l,s) ç N = N((G,l); (G,l')) if and only if (1,~) ~ N, that is wh en ( 1, :r )( G, 1) Ç ( G, 1 )( 1,s) • But ( 1, :S )( G, 1) = {(1, !)(a,l) : at. G} = {(f(! ,a,l), q>(:s ,a,l)): a E.. G} Ç (G,l)(l,s) when <{'(J,a,l) =s for e.ll a E.. G • Hence (G,~)/(G,l)~ ~ if and only if (4.6) holds. Finally, if (G,~)/(G,l) is a mixed group with nucleus {(G,l)(l,s) ; xE:. A} then we have the natural mapping onto ~ if and only if (G,l)(l,!)(G,l)(l,~) = (G,l)(l,I)(l,~) = (G,l)(l, ~(~,1,~)) is equal to (G,l)(l,s~) • This is simply (4.7). # Corollary 4.2 If S = (G, ~) is a group satisfying (4.1)- (4.7) then cp(l,b,~X.) ::sot when St~ for all b E: G, oc E= r . Proof q>(s,b,ol..) = (G,l) under the isomorphism a~ (a,l) (a é G) we have that S is an extension of G by the mixed group ~ • Conversely, it is ea~y to see that any extension of G by 1 is isomorphic to a skew product ( G, r ) satisfying (4.2} - (4.7). Definition 4.1 A skew product (G, ~) satisfying (4.1) - (4.7) is called an extension of the group G by the mixed group 1 wi th functions f, q:> • Equivalent extensions Definition 4.2 Two extensions H1 , H2 of a group G by a mixed group ~ with extension functions f1 ,~1 and f 2, ~2 are called equivalent if there is a G-isomorphiam (i.e. an isomorphiam leaving the elem- enta of (G,l) fixed) mapping H1 onto H2 (cf. Definition 2.3). In general, if Hi has elements (a,~)i (i = 1,2) then H1 and H2 are equivalent when there is an isomorphism ~ : H1 onto H2 auch that (4.8) (a,l)JF = (a.,l) 2 , (l,o<.)1-c = (xc(,:roC)2 where 01. ~ x is a.n arbi trary mapping of into G o( r and o/.. ~ ~«. ia a permutation of r . Theorem 4.3 Let Hl' H2 be extensions of G by with functions f 1' Cfl and f 2' permutation ()(-"'fa(. of r and a mapping ol --'1 xal of r into G such that (4.9) X / (sa(, bx , S\3) f (o(, b, f3)X Froof Suppose that H1 and H2 are equivalent then there is an isomorphism ~ : H onto H satisfying 1 2 (4.8). Then {(a,ot ) ( b, {(a,l) (1,1){) (b,l) (1,13) ~ -r: 1 ~) 1~-c = 1 1 1 1 = (a,l) 2 (x~,sa() 2 (b,l) 2 (x~,s~) 2 ( ax,, seL) ( bx ?>, "f!!) = 2 1 2 = (ax~f 2 (!,, bxt>' ~~), ~~("'i..(, bxf>, s~) ) 2 • But we also have {(a ot) (b, (3) }-r {( af (ot, bt(~), 9 1 1 = 1 = ( af1 ( ol ' b' ~ ) x q> t ( ~ ' b' ~) ' ~ ~1 ( o( ' b' ~) ) 2 • Therefore, because ~ is an isomorphism (4.9) and (4.10) must follow. Conversely, for any permutation 0{ -- s r1.. of r and tunction XO(. of l' on G a mapping ""C' defined by "t" : (a,0() ~ (axo(,s,.() is an ieomorphiem 1 2 when (4.9) aad (4.10) hold. This proves the theorem. # Corollary 4.3 The mapping Cl(~ s"'- is an automorphism of the mixed group -r (i.e. an isomorphism of r onto itself. See Definition 1.3). Proof Since (1,1)1-c = (1,1) 2 , s, = 1 , x1 = 1 • Renee (4.10) gives ~ (s~,b,l) . :s If O<.f: b. 2 = q>l (Ol, b, 1) • then Cf>i_(!"', b,l) =""fol for all b ~ G and therefore s~é~ (applying (4.6) twice). Conversely sK E: ~ implies r:J..f.~ and hence ~~I"" maps ~ onto itself. To prove the mapping is an automorphism we must show Iot "Ill = Sc~.~ ( o{, ~ E: 1) • This is sufficient since the mapping is one-one. But from (4.10) q> (sa( ,x , sf=l) = 2 10 and ao if ex e: f:::. then by corollary 4. 2 s 'Pt (0(. ' 1 ' ~) s~-s~ = ~c(.~ proving the assertion. # Definition 4.3 Two extensions H1 , H2 of a group G by a mixed group r are bi-equivalent if there is a H H G-isomorphism between 1 and 2 mapping the coset (G,l)L(l,~) 1 of H1 onto the eoset (G,l) 2 (1,~) 2 of H ( all r:J.. E: 2 r ) . 4.4 Two extensions H and H of a group Theorem 1 2 G by a mixed group with functions f , cp and r 1 1 f 2 ,~2 respectively are biequivalent if and only if there is a function xc(. of r into G auch that -1 -1 ( ) (4.11) t 2 Note If two extensions H1 , H2 of G by ~ are equivalent under an isomorphism 'L : (a,~) 1 __, (ax"',so() 2 we can wri te "t: = -c,-cl.. where "t:,: (a,o<.) ~ (axo~..,O{.) 1 3 and 1:: : (a,()() ~ (a, s.J • Then -r, is an isomorph 4 3 2 ism between extensions and T~ a G-isomorphism which is also a î-automorphism. 5. FARTICULAR EXTENSIONS Definition 5.1 An extension H of a group G by a mixed group r is normal when the mixed group r is a group. In the case of a normal extension the nucleus ~ = r and it follows from Corollary 4.2 that (5.1) q> («, b, f>) = o{t3 (all ai,~ E; l' , b E.. G) • It is clear an extension H = (G, ~) is normal if and only if ( G, 1) <1 H and the quotient group H/( G, 1) ~ r . We shall write f(~,b,l) = ~b • Then for a normal extension f(o(,b, ~) = f(oc, bf(l,l,(3), = f(c<,b,l)f(~(o<.,b,l), 1,(3) = a~~f(ol,l,~) using (4.3) and (5.1). The nontrivial ~onditions which remain among (4.2) - (4.7) may then be written (5.2) f(l,a,f>) = a (5.3) (a) 01! aoL,l,f3)f(oLt'3,1,"'6) = ~f(~,1,1)f(ol.,l,f3l) • The product (4.1) may be written (5.4) (a,o()(b,~) = (af(«-,b,(S),oq9) = (a~bf(o<.,l,f3),o(~). We note that (1,~)(a,1)(1,~)-1 = (f(~,a,l),~)(1,~)-l = ( f (~,a,l),l)(l,~)(l,~)-1 = (f(~,a, l ),l) so the map- at ping ~ : a~ a is an automorphism of G • From (5.3b) we note ~OZ and o<.p di ffer only by an inner auto- morphism (viz. that induced by f(oc,l,~)) and so the mapping ~ : rJ... ~ ~ is a mapping of r into the group of automorphisms A(G) of G auch that $ induces an antibomomorphism of r into the quotient group of cosets: of A(G) modulo J(G) (where J(G) 4 A(G) is the group of inner automorphisme of G ). This quotient group is called the group of outer automorEhisme (Zaes enhaus [1] p. 48) • Definition 5.2 A mapping 6 of a group r into the the group A(G) of automorphisme of a group G is called admissible if it induces an antihomomorphism of r into the group A(G)/J(G) of outer automorphisme. Two admis si ble mappings ~ : r:J.. ~ oc , for . - _, the sa.me r and G are called eJ.milar when ()(. , aL are in the sa.me co set (mod J ( G) ) for each 0( E: ï'. Lemma 5.1 If H = (G, ~) is a normal extension of G by the group r then the mapping e : o<. ~ ëi ( ot. é r) where ~: a _,.,<>La= f(oL,a,l) (a ~ G) is an admissible mapping. Biequivalent normal extensions have similar admissible mappings. Proof The first part follows from the discussion above. To prove the second part let H, H' b~ two auch normal extensions which are biequivalent and have functions f and f' respectively. Then by Theorem 4.4 we ca.n choose a mapping o(.-- xot. of r into G auch that f(~,b,l) = x:1 f·(~,b,l)x~ taking into aeeount that the exteneions are normal. Renee the mappings b ~ f(~,b,l) and b ~ f'(~,b,l) are two automorphisme of G differing by an inner automorphism and hence the corresponding admissible mappings are similar. # Because of Lemma 5.1 we may consider the set of normal extensions whose admissible mappings are sim ilar to a given admissible mapping 6. For a given G and given Ï there is no general guarantee that any such extension exista (see Baer (2] ). However when G is abelian we have : Lemma 5.2 There is always at least one normal exten sion H of an abelian group G by a group r whose admissible map is the same as a given admissible mapping e • Proof For an abelian group G the group J(G) of inner automorphisme is the unit group. Renee 6 is an antihomomorphism of r into A(G) and (5.3b) be cornes ~(~a)= ~~a • We therefore can define f(o<.,b,~) = Q(bf(o<.,l,j.?) = Olb and it is readily verified that this function satisfies (5.2) and (5.3) and hence gives the required normal ex tension. Definition 5.3 An extension H of a group G by a mixed group r wi th functions f, Cf is called a splitting extension when f(o<.,l,~) = 1 (all o(,{3 E.r) • Note that the exteneion of Lemma 5.2 is a ep1it ting extension. Theorem 5.1 An extension H = (G, ~) is a sp1itting extension if and on1y if (l,r) is a subgroup of H • Then H = (G,l)(l,r) expresses H as a product of two subgroups with unit intersection. Pro of A. If ( 1 tl') is a group then (1,~)(1,~) € (1,~ for al1 eX,~ é r and so (1,«-)(1, (0) = (f(~,1,~),~(~,1,~)) implies that f(o(,l,~) = 1 • B, Conversely, if F(o<.,l,j3) = 1 for all o(, ~ ~ 1 then (1,1) is a group since From Theorem 5.1 we see that the definition of sp1itting extensions generalises the concept of split ting extensions of normal group extension theory (see Kurosh tl] §52) and the notion of re trac ting factor systems of Baer ~] (see Zassenhaus [1] p. 128) • It is similar to the idea of factorisation of groups into permutable subgroups as ie eeen by the following theorem: Theorem 5.2 An extension H = (G, r) is biequival- ent to a splitting extension if and only if there is a subgroup U~ H such that H = (G,l)U and (G,l) nU= (1). Proof A. If H = (G,1)U and ~,1) n U = (1) then (U:1) = (r:1) and since U must have at least one representative from each coset (G,1)(l,o<.) (olE: r) , it contains exactly one. We write U = {(x~,oc) : ~éT'}. Because U is a group (xol,o<.)(x (xo<.f( B. If H is biequivalent to a sp1itting exten- sion then there is a function xol of r into G such that = 1 It is easily shown xcL f(oc, x 13 , \0) x Lemma 5.3 A necessary and sufficient condition that an extension H of a group G by a mixed group "l' is a sp1itting extension is that f(~,a,~) = f(oc,a,1) =~a • Proof If f(ol ,a, (0) = f(ot,a,l) then f(ol ,1,j3) = 1 by Coro11ary 4.1. Converee1y, if f(ot.,1,f->) = 1 for a11 ~J~E:-r then f(~,a,~) = f(~,af ( 1 ,1,~),~(1,1,~)) = f(~,a,1) f (~(~,a,1), 1 ,~) and so f(ot,a, f=>) = f(oc.,a,l) • # 40 If we write ~(o<..,l, ~) = oc · o~ a cr(ot,a,l) = ()(. then = oc.a o ~ using ( 4. 4). We can then write the conditions (4.1) - (4.7) as 0(. b (5,6) (a, ol)( b, ~?>) = (a b, Cl o ~) . 1 ' (5.7) (a) a= a ; (b) la= 1 ., (c) lo\3 =~ ., w. o(b o( (5,8) (a) b c = at( be) (b) 1 = 0( ; (c) o ( c ) (oc: b) c = ( 5.10) For all Il( E:. r there is a unique two-sided 1 -1 -1 inverse ( Theorem 5.1) 0(.- auch that r:J... o ot. = ()(. o 0' = 1 • (5.11) "!b = s for all b ~ G if and only if s~l:::.. ; ( 5.12) ~ o ot = set when sE:.~ • Theorem 5.3 In a splitting extension H of a group G by a mixed group f"' the elements of r form a group r: under the product oc o ~ = q>(a(, 1, j3) • This mul tiplic ation is an extension of the product previously defined in the mixed group r . Pro of T: is a group from Theorem 5.1. The last as- sertion of the theorem follows from (5.12). Lemma 5.3 If H is a normal extension of a Sylow p-group P then H is biequivalent to a splitting extension. Proof The lemma is trivial for P = (1) • We proceed by induction and suppose it is proved for all p-groups with order less than that of P • Let H be a normal extension of p by r ( p l (r: )) ) wi th function f • Let P' be the derived group of P • We define the transfer of H into P (cf. Zass enhaus [~ p. 167) as the mapping ~ : H into P/P' (a,ot)-c = n f(À,a,C() (mod P' ) • ~H' Because f(~,a,~)f(~~,b,~) = f(~,af(~,b, ~),~~) it fol- lowe easily that (a,~)~(b,~)-c = (af(~,b,~),~p)~ = {(a,oc)(b,~)~~ because of the commutivity in P/P' • Therefore ~ is a homomorphism of H into P/P' • We now show that the image of ~ is a non-trivial group. It is well-known that p' f. p so taking a e: P, a ~ P' we have (a, 1 )"t = n f(b.,a,l) (mod p') Aér • We wish to show that n f(À,a,l) ~ P' • In fact, AéT' because a~ P' the automorphic image f(À,a,l) ~ P' since P' is a characteri~tic subgroup. However, p t Therefore ker 1: = K f. H and K t> P n K f. P • By the induction hypothesis Theorem 5.2 implies there exista U ~ K auch that K = U(P " K) and p l (U:l). Since K/(P n K) ~ KP/P = H/P , (H:P) = (K: P ~ K) and so H = UP (P ~ U = (1)) • This proves the theorem. # Theorem 5.4 If H = (G, r ) is a normal extension of a group G by a group r and ((G:l),(P:l)) = 1 then H ie biequivalent to a sp1itting exteneion. Pro of For G = (1) the theorem is trivial. We proc- eed by induction and suppose the theo rem is proved for a.l1 groups G* with order lees than the order of G • Let p be a Sylow p-group of G • Th en p is also a. Sylow p-group of H and then by the Second -1 Sy1ow Theorem all the Sylow p-groups x Px (x ~ H) conjuga.te to P in H are conjugate to p in G. -1 Bence x Px = for eome u é G • Thus 1 xu- E: N(P;H) and writing N = N(P;H) we therefore have x ~ NG (all x é H) and so NG = H. (cf. the proof of Theorem 3.2). We now have two cases: (i) N = H • Then P ~ H and so by Lemma 5.3 there is V ' H auch that H = PV and P ~ V = (1) • Then G n V V = U(G ~V) with U ~ (G "V) = (1) for some U ~V • ( Theorem 5. 2). Fina.11y, we note that (( G: 1), ( U: 1)) = 1 and so H = UG with Ur. G = (1) and H is therefore biequivalent to a sp1itting extension. (ii) N ~ H. Then we may apply the induction hyp- othe sis to N as a normal extension of N ~ G wi th N/(N A G) ~ NG/G = H/G • So N = U(N r. G) with N A G " U = G A U = ( 1) • But ( U: 1 ) = (N; N " G) = (H;G) and ao since U " G = (1), H = GU • This proves the theorem. # 6. STRUCTURE OF NORMAL EXTENSIONS Threughout this section we shall be considering normaL extensions. Let H be a normal extension of a group G by a group r with function f • From Lemma 5.1 H defines an admissible mapping of 1 into A(G) , viz. f3 : ()(~OZ where oc : a~cxa = f(oc,a,l) (a ~ G) • We shall consider extensions of G by ~ with admissible mappings similar to a given admissible mapping. Let r be a given group, Z a given abelian group and e z a given admissible mapping (and hence by the proof of Lemma 5.2 an antihomorphism) of r' into A(Z) • We then define as kernels all pairs ( G; e) of groups G wi th centre Z( G) ~ Z and admiss ible mappings e of r into A(G) auch that for all oc e. T' the automorphism cx.e : a~ 0(. a induces an auto morphism cxejz on the characteristic subgroup z such that oce\z = ~ez • We identify two kernels (G;e) = ( G' ; 6' ) when G ~ G' and under some isomorphism o.' between these groups e and u are similar. The set of all kernels we denote by L. k • As was noted in section 5 not all kernels have extensions and the subset of kernels (G;5) which do have extensions H of G by 1 will be denoted by L' e • we note that all along 1, z and e z are taken to be fixed. We denote the set of all extensions H of extendi ble kernels by L. a.nd identify two exten sions when they are equivalent extensions of ident- ified kernels. We now place a structure on :[ by defining the product of two extensions following Eilenberg and Mac Lane [2] as follows: Let Hi be an extension of (Gi;ei) (i = 1,2) • Then form the set W = \(~,a ,~) : ai ~ G., D(é-r} 2 l. with the product (6.1) (al,a2,~)(bl,b2'~) = (al f 1 ( 0(.' br ~)' a2f 2 ( o<..' b 2' ~)' o<. ~) where f , f are the functions of the extensions H , 1 2 1 H2 • It is easily verified that W is a group and in fact a subgroup of the direct product H1 x H2 • Fur thermore, G1 x G2 = i(a1 ,a2,1) ai 6 Gi} is clearly 1 a normal subgroup of W and N = [(z,z- ,1) : z ~ Z} is a characteristic subgroup of G x G • Therefore 1 2 N (6.2) a . E. G . , o<. c; T} = W/N • l. l. Lemma 6.1 H ® H €;: i.e. H ~ H is homomorphie 1 2 L 1 2 to ~ with kernel whose centre is isomorphic to Z • H H Proof 1 ® 2 is homomorphie to -r under the mapping N(a1 ,a 2 ,~) ~ oc The kernel of the homomorphism may be written (6.3) G d G {N(a ,a ,1) : ait. Gi) (G ><. G )/N 1 2 = 1 2 = 1 2 It is easily verified that Z(G o G ) tN(z ,z ,1) : 1 2 = 1 2 zi " Z} which is isomorphic to Z under the mapping N(z1 ,z2,1) = N(z1z2,1,1) -+ z1z2 • Finally we note that the associated admissible map ping of l' into A(G1 o G2) is G1o92 defined by (6.4) oe.(ê ): (a ,a ,1) ~ (o<.a ,0(a ,1) 1oe2 1 2 1 2 = (f1 (~,a1 ,1), f 2 (~,a 2 ,1), 1) • # Corollary 6.1 We can define the product of two kernele using (6.3) and (6.4) (Gl;el) o (G2;e2) = (Gl o G2; el 0 e2> E:. L:k • It is easily shown that the products in L and L.k are both associative and commutative. Furthermore we have: Theorem 6.1 L, L'k and Le are commutative semi- groups under the products defined by (6.2) and Corol lary 6 •.1. Furthermore L. is cancellative in the following sense: I:f" Hl' H2, H3 (:. L and have kernels (~1; el)' and (G ;e ) = (G ;e ) respectively then H ® H H ® H 2 2 3 3 1 2 = 1 3 implies that H H • 2 = 3 Proof To prove the first part it only remains to show that 2: e is closed under the given product. But if ( G ; ), ( G ; ) E. have 1 e1 2 e2 Le extensions H1 and H2 respectively, then (G1 ;e1 ) o (G 2 ;e 2 )~ ~e since thie kernel has the extension H1 ~ H2 (see the proof of Lemma 6.1). To prove the second part of the theorem suppose that Hl' H2' H3 E: L have functions fl' f2' f3 re- spectively. By the previous lemma Hl® Hi (i = 1,2) (G o G ; o ) are extensions of the kernel 1 2 e1 e2 = ( G G ; o 6 ) and we may suppose that the ir ex 1 ° 3 61 3 tension functions are fi (i = 1,2) • Writing the elements of G1 o Gi as N(a1 ,ai,l) = N(a1,ai) (cf. é 1, (6.3)) we have fi(o(, N{b1 ,bi), ~) (oc:,~ N(b1 ,bi) ~ G1 o Gi) as our functions. Since N{a1 ,ai,oc)N{b 1 ,bi'~) = N{a1 f 1 (~,b 1 ,~), aifi{~,bi'~), ~~) = (N{a1 ,a 1 )N{f1 (~,b 1 ,~),fi(~,bi'~)), ~~) it follows that i f (o<., N(b1 ,bi), ~) = N ( f l(Cl~. , bl , f! ) , fi( 1 into G1 ° G2 (with x1=y1=1) such that 2 N(x<>',yot)f (:Foe., N(b1 , b2)N(xll,yF- ), sp) 3 = f (o(, N(b1 , b 2 ), ~ )N(xoL(3 ,y N (x Ol f 1 ( :roL , b 1x ~ , "".f~) , y cl. f 2 ( !"o(. , b 2y (3 , s ta ) ) = N(fl (oc:, b 1 ,f3)XOL~' t 2(o(, b2, f3)Yotp) and therefore there exista z 6 Z auch that (6.6) -1 ( (6.7) y~f 2 ( Y~,b 2y~,s~) = z f 3 ~,b 2 ,~)y~~ • From (6.6) and (6.7) z is independant of b2 and b1 respectively so we can write z = z(~,~) and it follows immediately from (6.6) that z(~,l) = z(l,p) = 1 • But from condition (4.3) on f 1 we also get from (6~) (6.8) z(oc,~)z(~~,~) = z(~,~~) and so for « = 1 , Z(f->,1) = 1 for all ~,1 ~ -r. Renee from (6.7) y~f 2 (s~,b 2y~,Ip) = f 3 (~,b 2 ,~>Yœ~ which is the condition that H and H are equival 2 3 ent extensions of the kernel (G o G ; o 6 ) • Hence 1 2 e1 2 H2 = H3 • # Lemma 6.2 The number of extensions of a given kernel is finite. Proof The number of extensions is certainly no more than the number of possible functions f • Theorem 6.2 All extendible kernels ( G ; & ) ç 2: have e the same number of extensions. Proof rt· is sufficient to prove a one-one correspond ance between the extensions H of (G; e) E. L. and e the extensions of the (extendible) kernel (Z;6z) (see Lemma 5.2). In particular, if Ei (i = 1, •• m) is a complete set of extensions of ( z; ez) (finite by the previ ous lemma) and H is an extension of (G;e) then we show that H ® E. l. (i = 1, •. m) is a complete set of extensions of (G; ê) • First1y, if H is an extension of ( G; e) and E an extension of (z;ez) then H®E is an extension of the kernel (G o z; e o ez) by Lemma 6.1. But G o z !'!:! G under the mapping N(a,z) = N(az,l) __,.. az (a c: G, z ~ Z) and since the images of 8 and 9z a~t ident- ically on Z , (Go Z; So9z) = (G;e) and H ® E is an extension of (G;6) • Applying Theorem 6.1 we have that H ® Ei (i = 1, •• m) is a set of m dis tinct extensions of (G;e) . So (G;e) has at least m extensions. Secondly, we show that any extension H' of (G; e) may be written H' = H ® E for some extension E of (z;ez) • Let f and f' be the functions for H and H' respectively. Since both have the same kernel and in particu1ar the same associated admissible mapping e, equation (5.3b) holds for each with the same values for «(Pa) and ~~a • Therefore 1 f ' ( ()(.~a ranges over G as a does. We define z(~,l,~) €- z by z(o<.,l,~)f(O(,l,f3) = f' (ot,l, (.?.>) and .in general write z(o<.,a,~) = Olaz(oc,l,~) (a. cs Z) • Then beca.ul!le f and f' satisfy (4.2) and (4.3) it can be shown that z(oc,a,~) must also satisfy these conditions and hence z(~,a,~) is a function for an extension E of Fina1ly, H ®E = {N(a,z,o<.): a E:. G, z ~ Z, ~E:T""} is isomorphic to H1 by the mapping N(a,z,~) = N(az,l,~) _, ( az, IX) and so H 1 = H ® E • Thus the extensions of the kerne1 (G;6) are pre cisely H ® E1 (i = 1, •• m) and so the theorem is proved. # Theorem 6.3 The set of a11 normal extensions of the kerne1 (Z; ez) is an.- .abe1ian group. Proof Let S = tEi ; i = 1,2, •• m1 be the set of auch extensions. The number is finite by Lemma 6.2. We note that Ei ® Ej is a1so in S (cf. proof of Theorem 6.2) and so S is a commutative semigroup. Furthermore, by Theorem 6.1 S is a cance11ative semigroup and eince S ie finite it is therefore a group. # Corollary 6.3 The unit element of S is the splitting extension (cf. Lemma 5.2). Pro of The eplitting extension of sion E with f(o(,b,{3) = o(b • Let El be any other extension with function f' • Th en E ® E • = {.N(a,a',D() .• a, a• E. Z, o(. E. r} ~ E 1 under the mapping N(a,a 1 ,~) = N(aa•,1,~) ~ (aa•,~) as is easily verified. # Theorem 6.4 If Z = (1) each kernel (G;S) ~ ~k has exact1y one extension. Proof Since Z = (1) there ie only one extension of (Z;6z), viz. the splitting extension f(~,l,~) = 1 (~,~ ~ r) . Renee by Theorem 6.2 every extendible ker nel (G; S) has exactly one extension. It remains to show that every kernel is extendible •. Since the admissible mapping 6 defines auto morphisme ~ : a ____,«.a ( o<.. e.. T' , a e. G) for which ~ OC: and ~~ differ only by an inner automorphism and since for a group G with unit centre ea.ch inner automorph- ism is induced by exactly one element of G we can de- fine f(oc.,l,f~) as the unique element of G such tha.t oc:(fO a) = f ( oc,l,p ) oc:f3 af ( ~,l,p). -1 (cf. (5.3b)) and in particular f(l,l,~) = 1 • Then oc ( ~ ( "d a ) ) = o(. ( f ( ~ , 1 , o ) ~t af ( p , 1 , l ) -l ) 1 = ot f ( ~ , 1 , o ) oc: ( ~ a ) otf ( j3 , 1 , 1 ) -l = O(f(~,l,~ )f(oc.,l,~l)~ 6 a.f(ol,l,f>"'>-l f('J?>,l,"'6)-1 ; while, on the other hand, 1 0( ( 1\ a )) = f (aL , 1 , ~ ) o<:f3 (t a ) f ( o( , 1 , 13 ) -l = f(~,l,~)f(~~,l,1)0(~1 af(~F,l,l)- 1 f(O(,l,~)-l • Therefore, since the element of G inducing the inner 1 automorphism fil.. (f! (' a)) --'Jo ""P a is unique, i t followe that f satisfies (5.3c). Renee defining f(oc,b,~) = otbf(oc,l,~) we have an extension function eatisfying conditions (5.2) and (5.3). This gives an extension of the kernel (G;e) and proves the theorem. # Lemma 6. 3 For any kernel ( G; 6) ~ L.k the re is a kernel (~; è) E. L.k auch that (G; 8) o (G; ë) is ex tendible. Proof We define ~ = {a : a E:. G} with ao = Da and z = z ( z é Z) • Similarly if oLS ; a --oc. a ( 0( é T', a ~ G) then we define e by o(e : a_, ~a . It is easily seen that ë is also an admissible mapping. We now wish to find f((l(, N(a, o), ~) (0(,{3 el', N(a, b) ~ GoG) such tha.t f satisfies (5.2) and (5.3) and hence defines an extension of the kernel ( G 0 G; e 0 e) by ,..., • We 1 note that N(z,z- ) = N (z é Z) is the unit element of G o ~ • Since 6 is an admis si ble mapping, (,e) (!11'.8) and <~we differ only by an inner automorphism and so we can find elements cae,~ E:. G ( 0( , ~ ~ l' ) such that &c: (~ a.) = c ~~ac-l for all a é G (cf. (5.3b)). In partie- «.}J3 0(,~ ular, we may take c., = c ~ = 1 • By comparing the .... ,, 1.~,- inner automorphiems mapping ~(~( 1 a)) onto ~~1 a (cf. the proof of Theorem 6.4) we find (6.9) col,f3cal. , E. • 15 1 Z~cf>;tcOl,~~ We now define f(o(, N(a,'6), ~) = N( f ( <>< , N, r) f (aL r , N, 1' ) f ( ()( , N, (6 1) -l aif (p., N, "6' ) -l --1 ) ( --1 ) ( -1 - ) ( o( -1 - ) = N ( col,f!, cot,(3 N catp,J' , cocf?>,l' N c~,(3'a' , c0t',f3'a' N cp.,-t , "c~,l' 1 = N(z,z- ) for some z é z , by (6.9) = N (the unit element) • Thus f is a suitable extension function and (G;e) o (~;ë) is an extendible kernel. # A nilpotent group may be de.fined (cf. Kurosh [1] §62) as a group which has a non-trivial centre for each non- trivial quotient group. It is easily shown that each finite group G has a unique maximal nilpotent normal subgroup M(G) and that the corresponding quotient group G/M(G) has centre (1) • Furthermore, since M(G) is unique and maximal it is a characteristic subgroup of G • Using the resulta of Theorem 5.4 and 6.4 we can then prove: Theorem 6.5 A k ernel (G;e) has exactly one exten- sion by a group f1 in the case that the maximal nor mal nilpotent subgroup M(G) of G has order coprime to both (r:l) and (G: M(G)) • Proof Because M(G) is a characteristic subgroup of G we can define an admissible mapping e': r into A(G/M(G)) by oce' : Me.---+ M«a (aM t: G/M) when od~ : a ~Ka • Since G/M has centre (1) the kerne1 (G/M; 8') has exact1y one extension by the group ., (Theorem 6.4) • Ca11 this extension R. Since ((R:1),(M:1)) = 1 then if we define eM by oteM = (oLS)IM , t.he lœrne1 (M;eM) has exact1y one extension by the group H and we denote this extension by H • We now note that H is the unique extension of the kerne1 (G;6) by 1. C1early there can be no other extension and since G is the unique extension of (M;~) by G/M (Theorem 5.4) and H contains auch an extension H is the unique exten- sion of (G; e) • # 7. STRUCTURE OF GENERAL EXTENSIONS For general extensions we consider in place of the kernel of the normal extensions the u-group (G,K;B) , and in place of extension by a group we consider exten sion by a particular representation of a mixed group. Let B/A be a particular representation of a mi&ed group r ' z be an abelian group and e z an admissible mapping of B into A(Z) • We call (G,K;B) au-group if (K;S) is a kernel with centre Z for a normal extension by B and if G is a normal extension of the kernel (K;8) by A where 8 = elA (A~ B) • We de note the set of all auch u-groups by U identifying 1 (G,K;0) = (G',K';6 ) when (K;B) = (K•;e') (see sec tion 6) and G and G' are equivalent exteneions. Then using definitions (6.2), (6.3) and (6.4) we have ( 7 .1) ( G , K ; 8 ) o ( G , K ; 6 ) G ® G , K o K ; ~ o ) ..: 1 1 1 2 2 2 = ( 1 2 1 2 e2 It is immediately obvious from Theorem 6.1 that this makes lJ a commutative semigroup. Now considering extensions of these u-groups we call Han o-group of (G,K;e) if H is a normal extension of (K;e) by B auch that H ~G. Two o- groups are identified when they are equivalent extensions of the same kernel. In general not all u-groups have o-group extensions and we denote by lJe the subset of extend ible u-groups. It is easily proved (cf. Theorem 6.1) tha.t U e is a subsemigroup of U . Lemma 7.1 The group G- of normal extensions Fi (i = 1, •• m) of (Z;8z) by B has a subgroup ~ of those extensions which are o-groups for the u-group (E,Z;6z) where E is the splitting extension of (Z;ez) (cf. Theorem 6.3). The cosets of G modulo ~ are just the sets of extensions which are ô-groups of the same u-groups. Proof Becaus~ of Corollary 6.3 = (E,Z;6z) • Therefore two b-groups F, F' with u-group (E,Z;Sz) (i.e •. F,F' E. ~) have F ® F' E:. ~. Since G is finite this proves that ~ is a subgroup. It is clear that all the extensions of the coset Z. ® F have the same u-group as F do es. Conversely, if F, F' are o-groups for the same u-group (Ë,Z;8z) then, since G is a group, there is an o-group Fi with u-group (Ei,z;ez) auch that Fi® F = F' • Since the respective u-groups must satisfy Theorem 7.1 If (K;e) is an extendible kernel for B then the normal extensions Hi (i = 1, •• m) of (K;S) fall into classes such that two extensi ons in the s ame class are both o-groups for the same extendible u-group. Thesè classes all have equal numbers of elements. Proof The theorem follows from Lemma 7.1 since if Fi (i = 1, •• m) is the set of normal extensions of (Z; Sz~ by B then the set of normal extensions of (K;S) by B may be written H ®Fi (i = 1, •• m) for any extension H of (K;e) • (cf. the proof of Theor- em 6. 2). # Corollary 7.1 All extendible u-groups (G,K;e) have the same number of different o-groups as extensions. Proof The number of extensions is (~:1) where ~ is defined in Lemma 7.1. # Definition 7.1 are said to be similar if there are two extendible u-groups ) and ~ ,I ;ë ) such that (G1 ,K1 ;ê1 2 2 2 (Gl,Kl;'&l) o (~l,!l;ël) = (G2,K2; e2) o (112,R2;é2) • Note that similarity is an equivalence relation, i.e. it is symmetric, reflexive and transitive. We may there- fore define <(G,K;e)> to be the class of all kernels similar to (G,K;S) • It is easily seen to true that the product (7.2) ((G ,K ;e )>o((G ,K: ;ë )> = ((G ,K ;e ) o (G ,ï2.;e )> 1 1 1 1 1 1 2 2 2 .... 2 2 is well-defined and that this product turns the set of similarity classes into a semigroup with unit element U e • In fact, we have more: Theorem 7.2 The set of all similarity classes of u-groups forma a group under the product (7.2). Proof It remains to show that each class has an in- verse class, or, equivalently that for each u-group (G,K;e) there is au-group (G,K;ë) such that (G,K; e) o (G,K; e) ~ Ue • We define (K; é) as in Lemma 6.3 so that (K;e) o (K;ë) is an extendible kernel. The re is at least one u-group with kernel (K;e> ' viz. (~,fë; è) with ~ = {x . x e. G} su ch that 'XY= y x and z = z (z ~ Z) • Let ( Gi, K; ë) (i = 1, . . m) be the set of all u-groups with kernel (K; è) • Th en the u-groups ( G,K; e) o (Gi ,!; ë) (i = 1, •. m) forma com- plete set of u-groups with kernel (K;e) o (K;ë) by Theoreme 6.1 and 6.2. But this latter kernel has an extension H by the group B because of the way (K;ë) was chosen and H must have one (G,K;8) o (Gj,K;ë) as a corresponding u-group. Therefore (G,K;S) o (~j,l;ë) is extendible as required. Renee ((G,K;e))-l = {(Gj,K;S)) and the theorem is proved. # Theorem 7.3 If K is a group such that its maximal normal nilpotent subgroup M = M(K) (see the end of section 6) has order (M:I) coprime to both (B:l) and (K:M) , then au-group (G,K;6) has exactly one o-group H as an extension by a representation B/A of a mixed group. Pro of By Theorem 6.5 there is exactly one normal ex- tension H of the kernel (K; e) by B • This exten- sion must contain a subgroup G* which is an exten sion of (K;ë) by A (where ë = elA ). However, from Theorem 6.5 again, (K;&) has only one normal ex- tension by A and this extension is therefore G • Renee (G,K;9) has the unique o-group H. # Theorem 7.4 Let B/A be a representation of a mixed group r and let -,:;. be a homomorphism mapping G onto A with kernel K • Then G may be considered as an extension of a kernel (K;ë) by a group A • Furthermore, if K is a group such as is defined in Theorem 7.3 then the number of extensions H of G by B/A corresponding to ~ is the same as the num- ber of non-similar admissible mappings 9 : B into A(K) with elA = ë • Pro of Choose ua( ~ G su ch that u o(.""t :;0(. ( oc e: A) • -1 Th en we de fine e . A into A(K) by O(ë . a~ u"a ue<. (a ~ K) • It is clear that the admissible mapping ë is determined to within a similarity claas. Since equivalent extensions have similar mappings the theorem is then a restatement of Theorem 7.3. # 60 NOTES 1., Mixed Groups The concept of mixed group was first introduced by A. Loewy (Crelles Journal ~57 (1927) 239-254) but Definition 1.1, most of the theoreme of this section, and Theorem 2.1 are due to Baer [11 • Further refer- ences may be: found in Tschebotarow and Schwerdtfeger [l] p. 97-98. 2. Extension Problem Except for Definitions 2.5 and 2.6 and Theorem 2.1 the definitions and resulta of this section are new. In the counterexample given at the end of this section G is not extendible by B/A because the associated antihomomorphism of A into the automorphism classes of G is not extendible to B (see Theorem 7.5). This is not a necessary condition for non-extendibility. For a case where the associated antihomomorphism is extendible but G is not (in the case A = (1) ) see Baer [2] p. 415. ).. Cardinal Representations of Mpill The resulta of this section may be interpreted as a description of the transitive subgroups of degree p and ~s such appear to be new. 61 4. Extension Functions The idea behind this section is an application of the theory of skew product introduced by Redei '[1.] • However since the latter paper deals principally with products of pairs of groups there is little resemblance between the resulta. The resulta of section 5 look much more familiar. The part devoted to equivalent extensions may be compared with analagous resulta in the the ory of normal extensions ( e. g. Kurosh [l] § 48 equation (8)). 5~ Particular Extensions The resulta on normal extensions are well known. Theorem 5.4 is due to Schur and an alternative proof is given in Zassenhaus [~ p. 162. The concept of split ting extensions is closely related to the concept of factorisation of groups into permutable subgroups. This latter is discussed, for example, in Ore [1] • 6.. Structure of Normal Extensions The work of this section closely follows Eilenberg and MacLane [2] and Kurosh [:1] §51. (Note the slight difference in our definition of 8). Theorem 6.1 is an important tool in proving many of the resulta. Theor em 6. 4 was first proved by Ba er [2] p. 380. Lemma 6. 3 is a new proof of a lemma of Eilenberg and MacLane [~ and Theorem 6.5 is a new extension of these resulta made 62 particularly for application in the proof of Theorem 7.3, 7. Structure of General Extensions It is interesting that very many of the resulte for the normal case carry over into the general case. Def inition 7.1 and Theorem 7.2 generalise the section 6 of Eilenberg and MacLane ~]. Theorem 7.4 gives a criter ion sufficient for the extensions of a group G by a given representation of a mixed group. All the resulte of this section are new. BIBLIOGRAPHY Baer, R. L~ Zur Einordnung der Theorie der Misch gruppen in die Gruppentheorie. Sitz. Heidelberg (1928) Nr. 4. [~ Erweiterungen von Gruppen und ihren Iso morphismen. Math. Zeit. 38 (1934) 375-416. [3] Extension types of abe lian groups. American J. 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[~ Gruppen theorie (Springer-Ver1ag 1956). Tschebotarow, N. and Schwerdtfeger, H. (11 Grundzuge der Galois'schen Theorie (Noordhoff 1950). Turing, A.M. lü The extensions of a group. Compositio Math. 5 (1938) 357-367. Zassenhaus, H. [11 Theory of Groups (Chelsea 1958) 2nd Edition.