Scientiae Mathematicae Japonicae Online, Vol. 5, (2001), ON FRATTINI EXTENSIONS. Received May 2, 2001; revised June 28, 2001

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1 Scientiae Mathematicae Japonicae Online, Vol. 5, (2001), ON FRATTINI EXTENSIONS JAMSHID MOORI 1 AND B.G. RODRIGUES 2 Received May 2, 2001; revised June 28, 2001 Abstract. Let G be an extension of N by H. WesaythatG is a Frattini extension if N is a subgroup of the Frattini subgroup of G. In this paper we establish some properties of the Frattini extensions. 1. Introduction The motivation for this paper comes from an ongoing research by the authors on the theory of group extensions. In [2] B.Eick and H.Besche outlined an algorithm to determine, up to isomorphism, all finite soluble groups of a given order. This method which relays strongly on the theory of Frattini extensions is known as the Frattini Extension Method and it has been implemented in the computer algebra systems GAP and MAGMA. In the present paper we deal with theoretical aspects of Frattini extensions and our main results are outlined in Theorems [2.1, 2.3, 2.6, 2.11 and 2.14]. We also provide more detailed proofs to some known results and these are listed in Theorems [2.7, 2.8, 2.10 and 2.18]. In addition in Example 2.16 we determine the non-abelian groups of order 16 and 32 for which Φ(G) is afrattini extension of G 0 byφ(g=g 0 ): We encourage readers to consult [[7], [11], [12],[13]] for a background material on group extensions and on Frattini subgroups. We would like to thank Derek Holt for helpful comments. In particular he pointed out an article by R. L. Griess and P. Schmid (see [8] ) and the reference to Theorem The authors would also like to thank the referee whose suggestions and comments led to significant improvement of the content and the presentation of this article. 2. Frattini Extensions If N and H are two groups, then an extension of the group N by the group H is a group G having a normal subgroup K ο = N and G=K ο = H. Equivalently an extension can be defined in terms of short exact sequences of groups and homomorphisms as follows: Let ffi and ψ be the isomorphisms described above. Consider N! ffi K! i G and G! ß G/K! ψ H; where i is the inclusion map and ß is the natural homomorphism. Let ff = i ffi ffi and ffl = ψ ffi ß: Then we have the following short exact sequence f1 N g! N! ff G! ffl H! f1 H g; which wesay it represents the extension (G; ffl): Given a ( multiplicativelly written) group G; a G-module is an (additivelly written) abelian group A together with an action of G on A such that the following axioms hold for 2000 Mathematics Subject Classification. 20E22, 20D25. Key words and phrases. Group Extensions, Frattini Extensions, Frattini Subgroups. 1 Research grants from UNRF (University of Natal) and NRF (South Africa) acknowledged. 2 Postgraduate studies bursary from DAAD (Germany ) and the Ministry of Petroleum ( Angola) acknowledged.

2 122 Jamshid Moori and B.G. Rodrigues all a; b 2 A; g; h 2 G : (i) (a + b)g = ag + bg; (ii) a(gh) =(ag)h and (iii) a1 G = a: An extension (G; ffl) is called a Frattini extension if the Kernel of ffl is contained in the Frattini subgroup Φ(G) of G: Theorem 2.1 gives a useful characterization of the Frattini extensions for finite groups. Theorem 2.1. Let G be an extension of N by H and assume that G is finite. Then G is afrattini extension of N by H if and only if G=Φ(G) ο = H=Φ(H): Proof. Assume that G is a Frattini extension of N by H: Since N» Φ(G)and G=N ο = H; we haveφ(g)=n =Φ(G=N) ο = Φ(H): Thus H=Φ(H) ο = (G=N)=Φ(G=N) ο = (G=N)=(Φ(G)=N) ο = G=Φ(G): Conversely assume that G=Φ(G) ο = H=Φ(H): Then G=Φ(G) ο = (G=N)=Φ(G=N); since H ο = G=N: Since Φ(G=N) Φ(G)N=N ο = Φ(G)=N Φ(G); we have jgj jφ(g)j = jgj jnj 1 jφ(g=n)j» jgj jnj jn Φ(G)j jn Φ(G)j : jφ(g)j We deduce that 1» jn : Thus jn j»jn Φ(G)j and hence jn j = jn Φ(G)j: Therefore j N = N Φ(G); which implies that N» Φ(G): The following Lemma [Lemma 2.2] is a stated problem by D.Holt and W.Plesken in [9]. We give a detailed proof for this Lemma and derive some new properties of the Frattini extensions. Lemma 2.2. [9] Let (G; ffl) be an extension by H in which Ker(ffl) is a finite H-module. If L is a maximal subgroup of G which does not contain Ker(ffl); then (L; ff) is an extension with ff = fflj L and Ker(ff) a maximal H-submodule of Ker(ffl): Proof. Let f1 N g!n! G ffl! H! f1 H g be the given extension where N = Ker(ffl): We need to show that: (i) (L; ff) is an extension of L N by H: (ii)ker(ff) is a maximal H-submodule of Ker(ffl): (i) Since N 6» L and L is a maximal subgroup of G; L < LN» G: But L is maximal in G implies that LN = G: Now, since (G; ffl) is an extension we have that G=N ο = H; so LN=N = G=N ο = H: But LN=N ο = L=N L ο = H implies that L is an extension of N L by H: Also note that Ker(ff) = fl j l 2 L; ff(l) =1 H g = fl j l 2 L; ffl(l) =1 H g = fl j l 2 L; l 2 Ker(ffl)g = L Ker(ffl) =L N: (ii) Suppose to the contrary, that is, Ker(ff) =N L is not a maximal H-submodule of Ker(ffl): Then there exists an H-submodule K of N such that N L<K<N:Since N 6» L; we deduce that L < LK < G: This contradicts the maximality ofl in G: Hence N L is a maximal H-submodule of Ker(ffl): Now if Ker(ffl) is an irreducible H-module, we use Lemma 2.2 to prove the following result. Theorem 2.3. Let (G; ffl) be an extension by H. Suppose that Ker(ffl) is a non-trivial irreducible H-module. Then (G; ffl) is a Frattini extension if and only if it is non-split.

3 ON FRATTINI EXTENSIONS 123 Proof. Suppose that f1 N g!n! G! ffl H! f1 H g (Λ) is a non-split extension. We needto show thatn = Ker(ffl) is a subgroup of Φ(G): If N 6» Φ(G); then there exists L a maximal subgroup of G such that N 6» L; andsoby Lemma 2.2 we have that (L; ff) is an extension, with ff = fflj L : Moreover, Ker(ff) is a maximal H-submodule of Ker(ffl) and since Ker(ffl) is irreducible as an H-module we have thatker(ff) =N L = f1 G g or Ker(ff) =N: If Ker(ff) =N we have that N» L which isacontradiction. Hence Ker(ff) =f1 G g and therefore ff is a monomorphism. But ff is an epimorphism, since (L; ff) is an extension. Hence ff is an isomorphism, and so ff 1 : H! L is also an isomorphism. Since L» G; we have thatff 1 : H! G is a monomorphism. Now for all h 2 H we have (fflff 1 )(h) =ffl(ff 1 (h)) = ffl(l) where ff 1 (h) =l for some l 2 L = ff(l) =h: Hence (fflff 1 )=I H ; and so (*) splits, which is a contradiction. Conversely, suppose that (G; ffl) isafrattini extension. We need to show that (Λ) is non-split. Suppose that (Λ) splits. Then there exists a homomorphism say ffi from H into G such that fflffi = I H ; and so ffi is a monomorphism and therefore H ο = Im(ffi): Hence ffi : H! Im(ffi) is an isomorphism. Also, we have that Im(ffi)» G and Ker(ffl)» G: It is easy to show that G = Ker(ffl):Im(ffi) and Ker(ffl) Im(ffi) =f1 G g: Thus G = Ker(ffl):Im(ffi) is a split extension. Since Im(ffi) 6 Ker(ffl) thenim(ffi) is not a maximal subgroup of G: Thus there exists L a maximal subgroup of G such thatim(ffi) <L<G:But maximality of L implies that L Φ(G) Ker(ffl): Now, Ker(ffl)» L and Im(ffi) <Limplies that L Ker(ffl)Im(ffi) =G; which iscontradiction. Remark 2.4. A simplest example of a Frattini extension is a non-split extension with Ker(ffl) an irreducible H-module. Corollary 2.5. If (G; ffl) is a Frattini extension by H with Ker(ffl) 6= f1 G g; then (G; ffl) is non-split. Proof. The proof follows from the argument used in the second part of the proof of Theorem 2.3 or alternatively we can use Lemma 11.4 on p. 269 of [12]. Theorem 2.6. If (G; ffl) is a Frattini extension by H then (Φ(G); ffl) is an extension of Ker(ffl) by Φ(H): Proof. Since (G; ffl) is an extension, we have G=Ker(ffl) ο = H; and so Φ(G=Ker(ffl)) ο = Φ(H): Since Ker(ffl)» Φ(G); we have that Φ(G=Ker(ffl)) = Φ(G)=Ker(ffl): Theorem 2.7. [9] Let G be a finite group and let (G; ffl) be afrattini extension of N by H with H a perfect group. Then G is perfect. Proof. Suppose that G is not perfect. Set N = Ker(ffl): Thus we have that G=N ο = H: Now (G=N) 0 ο = H 0 = H imply that (G=N) 0 ο = G=N (Λ): Since G 0 N» G and G 0 N N; we have (G=N) 0 = G 0 N=N» G=N: So by (*) we have that G 0 N=N = G=N: Since G is finite and G 0 N» G; we have that jg 0 Nj = jgj: Hence G 0 N = G: Since N» Φ(G); there is no proper subgroup K of G such that KN = G: (See Lemma 11.4 on p. 269 of [12].) Hence G = G 0 and the result follows. In [3] J.Cossey,O.Këgel and L. Kovács stated the following results (Theorem 2.8, Proposition 2.9 and Theorem 2.10) about Frattini extensions to which we give detailed proofs.

4 124 Jamshid Moori and B.G. Rodrigues Theorem 2.8. [3] If ffl : L! M is an epimorphism of finite groups and G is minimal among the subgroups of L; with ffl(g) =M; then (G; fflj G ) is a Frattini extension. Proof. Let K be a maximal subgroup of G such that Ker(fflj G ) 6 K then we have K< K:Ker(fflj G )» G: Now maximality ofk implies that K:Ker(fflj G )=G: Also we have that Ker(ffl) L, Ker(ffl) Ker(fflj G ) and K» L: Thus K» G = K:Ker(fflj G )» K:Ker(ffl)» L (1): Let W = K:Ker(ffl); then ffl(w ) = fffl(kx) j k 2 K and x 2 Ker(ffl)g = fffl(k)ffl(x) j k 2 K and x 2 Ker(ffl)g = fffl(k) j k 2 Kg = ffl(k): Now by(1)wehavethatffl(g)» ffl(w )» ffl(l); that is M» ffl(w )» M: Thus ffl(w )=M and hence ffl(k) = M: Since K < G and ffl(k) = M; minimality of G produces a contradiction. It follows that Ker(fflj G ) is contained in every maximal subgroup of G and hence in the Frattini subgroup of G: Therefore (G; fflj G )isafrattini extension. Proposition 2.9. [3] If (G; ff) isafrattini extension by F, and fi : F! G is a homomorphism, such that fffi is surjective, then fi is surjective. Proof. Let K = Im(fi) = fi(f ): Since fffi is surjective, we have that ff(g) = F =(fffi)(f ) = ff(fi(f )) = ff(k): So for any g 2 G; there exists k 2 K such thatff(g) =ff(k): We have, ff(g) =ff(k) ) ff(gk 1 )=1 F ) gk 1 2 Ker(ff) ) g 2 Ker(ff)K; 8g 2 G ) G Ker(ff)K: (1) Now, K» G and Ker(ff) G imply that Ker(ff)K» G: Hence by (1) we have that G = Ker(ff)K: Since Ker(ff)» Φ(G); as in the proof of Theorem 2.7 we deduce that K = G: Hence fi is surjective. Theorem [3] Composites of Frattini extensions are Frattini extensions. Proof. Consider f1g!ker(ff)! G ff! H! f1g with ff an epimorphism and Ker(ff)» Φ(G) and let f1g!ker(fi)! H fi! M! f1g with fi an epimorphism and Ker(fi)» Φ(H): We needtoshow that fiff is an epimorphism and that Ker(fiff)» Φ(G): (i) fiff is an epimorphism: Since h 2 H; then there exists g 2 G such that ff(g) =h: If m 2 M; then there exists h 2 H suchthatfi(h) =m: Hence (fiff)(g) =fi(ff(g)) = fi(h) =m: (ii) Ker(fiff)» Φ(G) : Since ff(g) =H and ff is a homomorphism, we have ff(φ(g)) = Φ(ff(G)) = Φ(H): Thus Φ(G) =ff 1 (Φ(H)); the inverse image of Φ(H): But, Ker(fiff) = fg j (fiff)(g) =1 M ;g2 Gg = fg j fi(ff(g)) = 1 M ;g2 Gg = fg j ff(g) 2 Ker(fi) ;g2 Gg = fg j g 2 ff 1 (Ker(fi))g: Since Ker(fi)» Φ(H); wehavethatff 1 (Ker(fi))» ff 1 (Φ(H)) = Φ(G): Hence Ker(fiff)» Φ(G): The following theorem is a straightforward consequence of a theorem of Gaschutz, which states that if N is a finite group with Inn(N)» Φ(Aut(N)), then there exists a finite group G such that N G and N» Φ(G): The converse of this statement has been recently proved in [6]. Theorem Let N be a finite abelian group. (G; ffl) with Ker(ffl) =N: Then there exists a Frattini extension

5 ON FRATTINI EXTENSIONS 125 Proof. Since N is abelian, Inn(N) =f1 Aut(N) g and hence Inn(N)» Φ(Aut(N)): Thus there exists a finite group G with N G and N» Φ(G) : If ffl is the natural homomorphism from G into G=N; then Ker(ffl) = N: Hence (G; ffl) is a Frattini extension with Ker(ffl) = N: Remark If N is an elementary abelian group, then the following theorem (see [[5] or [8]]), due to Gaschutz, describes the existence of a Frattini extension G of N by H. Theorem Let p be a prime, and let H be agroup whose order is divisible by p. Then there exists a group G with a non-trivial normal elementary abelian p-subgroup N such that (i) N» Φ(G); and (ii) G=N ο = H: Proof. See Theorem 11.8 of [5]. Now ifg is a p-group we will show that Φ(G) is an extension of G 0 byφ(g=g 0 ). Theorem If G is a p-group, then Φ(G) is an extension of G 0 by Φ(G=G 0 ): Proof. Since G is a p-group, G is nilpotent and hence G 0» Φ(G): Also G 0 G implies that G 0 Φ(G): Since G 0» Φ(G); we have that Φ(G=G 0 )=Φ(G)=G 0 : We deduce that Φ(G) is an extension of G 0 byφ(g=g 0 ): Remark The extension given in Theorem 2.14 is not a Frattini extension in general. For example take G = Q; the group of quaternions or G = D 8 : It can be easily checked that G 0 =Φ(G) ο = C 2 and Φ(Φ(G)) = f1 G g: Thus G 0 6» Φ(Φ(G)) and hence the extension Φ(G) of G 0 byφ(g=g 0 )isnotafrattini extension. In the following we determine all the non-abelian groups of order 16 and 32 for which G 0» Φ(Φ(G)); respectively. Example From the library of small groups of GAP4 [14] we found that there are 14 non-isomorphic groups of order 16. Only 9 of these groups are non-abelian. We consider these non-abelian groups and label them as G i with 1» i» 9 as listed in the first row of Table 1. The remaining rows of Table 1 have been computed by using GAP. It can be observed from Table 1 that the group G 3 is the only non-abelian group of order 16 for which Φ(G) isafrattini extension of G 0 byφ(g=g 0 ): Table 1 : Non-abelian groups of order 16 G G 1 G 2 G 3 G 4 G 5 G 6 G 7 G 8 G 9 Φ(G) V 4 V 4 C 4 C 4 C 4 C 4 C 2 C 2 C 2 G 0 C 2 C 2 C 2 C 4 C 4 C 4 C 2 C 2 C 2 Φ(Φ(G)) f1 G g f1 G g C 2 C 2 C 2 C 2 f1 G g f1 G g f1 G g Similarly there are 51 non-isomorphic groups of order 32. Only 44 of these are nonabelian and are labeled as G i with 1» i» 44: These groups are listed in the first row of Table 2. As in Table 1, we completed the remaining rows of Table 2. We observe that there are 6 possible candidates among G i ; 1» i» 44 for which wemayhave G 0» Φ(Φ(G)); namely G 2 ;G 3 ;G 10 ;G 14 ;G 32 and G 33 : It is clear that G 14 ;G 32 and G 33 satisfy the condition G 0» Φ(Φ(G)): Further investigation eliminates G 3 and G 10 : Hence G 2 ;G 14 ;G 32 and G 33 are the only non-abelian groups of order 32 for which Φ(G) isafrattini extension of G 0 by Φ(G=G 0 ):

6 126 Jamshid Moori and B.G. Rodrigues Table 2 : Non-abelian groups of order 32 G G 1 G 2 G 3 G 4 G 5 G 6 G 7 G 8 G 9 Φ(G) C 2 3 C 4 C 2 C 4 C 2 C 2 3 C 4 C 2 C 4 C 2 C 4 C 2 C 4 C 2 C 4 C 2 G 0 C 2 C 2 C 2 V 4 V 4 V 4 C 4 C 4 C 4 Φ(Φ(G)) f1 G g C 2 C 2 f1 G g C 2 C 2 C 2 C 2 C 2 Table 2 : Non-abelian groups of order 32(continued) G G 10 G 11 G 12 G 13 G 14 G 15 G 16 G 17 G 18 G 19 G 20 Φ(G) C 4 C 2 C 4 C 2 C 4 C 2 C 4 C 2 C 8 C 8 C 8 C 8 V 4 V 4 V 4 G 0 C 2 C 4 C 4 C 4 C 2 C 8 C 8 C 8 C 2 C 2 C 2 Φ(Φ(G)) C 2 C 2 C 2 C 2 C 4 C 4 C 4 C 4 f1 G g f1 G g f1 G g Table 2 : Non-abelian groups of order 32(continued) G G 21 G 22 G 23 G 24 G 25 G 26 G 27 G 28 G 29 G 30 G 31 G 32 Φ(G) V 4 V 4 V 4 V 4 V 4 V 4 V 4 V 4 V 4 V 4 V 4 C 4 G 0 C 2 C 2 V 4 V 4 V 4 V 4 V 4 V 4 V 4 V 4 V 4 C 2 Φ(Φ(G)) f1 G g f1 G g f1 G g f1 G g f1 G g f1 G g f1 G g f1 G g f1 G g f1 G g f1 G g C 2 Table 2 : Non-abelian groups of order 32(continued) G G 33 G 34 G 35 G 36 G 37 G 38 G 39 G 40 G 41 G 42 G 43 G 44 Φ(G) C 4 C 4 C 4 C 4 C 4 C 4 C 4 C 2 C 2 C 2 C 2 C 2 G 0 C 2 C 4 C 4 C 4 C 4 C 4 C 4 C 2 C 2 C 2 C 2 C 2 Φ(Φ(G)) C 2 C 2 C 2 C 2 C 2 C 2 C 2 f1 G g f1 G g f1 G g f1 G g f1 G g As a concluding remark we would like to prove a result on minimal Frattini extensions. For further properties of the minimal Frattini extensions, readers are referred to [2]. A group G is called a minimal Frattini extension of N by H; if it is a Frattini extension of N by H and N is a minimal normal subgroup of G: Lemma Let N be a minimal normal subgroup of G: If N has a complement in G; then N Φ(G) =f1 G g: Proof. If N is complemented in G then G = NH and N H = f1 G g for some proper subgroup H of G: Since N G and Φ(G) G; we have N Φ(G) G: Now N Φ(G)» N and minimalityofn implies that N Φ(G) =f1 G g or N Φ(G) =N: If N Φ(G) =N then N Φ(G) and hence NH Φ(G)H; so that G =Φ(G)H: But this implies that H = G which is not possible, since N Φ(G). Thus N Φ(G) =f1 G g as required. We now use Lemma 2.17 to give a more detailed proof for the following theorem which was stated as Lemma 4.6 by Eick in[2]. Theorem Let G be an extension of N by H; where N is abelian. Then G is a minimal Frattini extension of N; if and only if N is a minimal non-complemented normal subgroup of G: Proof. Let N be a minimal non-complemented normal subgroup of G: If N 6» Φ(G); then there exists a maximal subgroup M of G such that N 6» M: Now N G and M» G; implies

7 ON FRATTINI EXTENSIONS 127 that M<NM» G: The maximality ofm implies that NM = G: Since M N N and N M M we have that N M NM = G: The minimality ofn in G implies that N M = f1 G g and hence M is a complement ofn; which is a contradiction. So N» Φ(G) and therefore G is a minimal Frattini extension. Conversely, if N is a complemented minimal normal subgroup of G; then by Lemma 2.17 we have that N Φ(G) =f1 G g and hence N is not a subgroup of Φ(G): Therefore G is not a minimal Frattini extension. References [1] H. U. Besche and B. Eick, The groups of order at most 1000, except 512 and 768, J.Symb. Comput. 27(4) (1999), [2] H. U. Besche and B. Eick, Construction of finite groups, J.Symb. Comput. 27(4) (1999), [3] J. Cossey, O. H. Këgel and L. G. Kovács, Maximal Frattini extensions, Archiv Der Mathematik 35 (1980), [4] M. Deaconescu, Frattini-like Subgroups of Finite Groups, Mathematical Reports, Harward Academic publishers GmbH, London, 1986, Vol 2, [5] K. Doerk and T. Hawkes, Finite Soluble Groups, Walter de Gruyter, Berlin, [6] B. Eick, The converse of a theorem of W. Gaschutz on Frattini subgroups, Math Z. 224 (1997), [7] D. Gorenstein, Finite Groups, Harper and Row Publishers, New York, [8] R. L. Griess, and P. Schmid, The Frattini module, Archiv Der Mathematik 30 (1978), [9] D. F. Holt and W. Plesken, Perfect Groups, Oxford University Press, New York, [10] B. Huppert, Eindliche Gruppen I, Springer-Verlag, Berlin, Heidelberg, [11] B. G. Rodrigues, On the Theory and Examples of Group Extensions, MSc Thesis, University of Natal, Pietermaritzburg, [12] J. S. Rose, A Course on Group Theory, Dover Publications, Inc., New York, [13] J.J. Rotman, An Introduction to the Theory of Groups, 4th Edition, Springer-Verlag, New York, Inc., [14] The GAP Group, GAP Groups, Algorithms, and Programming, Version 4.2; Aachen, St Andrews, 1999 ( University of Natal, Private Bag X01, Scottsville, Pietermaritzburg,, 3209, South Africa address: moori@nu.ac.za, rodrigues@nu.ac.za