7. Minimal Normal Subgroups and the Socle Throughout This Section G Is A
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34 NICK GILL 7. Minimal normal subgroups and the socle Throughout this sectionG is a nontrivial group. A minimal normal subgroup ofG is a normal subgroup K= 1 ofG which does not contain any other nontrivial normal subgroup ofG. The socle, soc(G) is the subgroup� generated by the set of all minimal normal subgroups (ifG has no minimal normal subgroups, then we set soc(G) = 1). (E7.1)Find the socle ofD 10,A 4,S 4,S 4 Z. × (E7.2)Give an example of a group that has no minimal normal subgroup. (E7.3)IfG is the direct product of a (possibly infinite) number of finite simple groups, then what is soc(G)? (E7.4)Give a characteri ation of almost simple groups in terms of their socle. Clearly soc(G) is a characteristic subgroup ofG. 26 Theorem 7.1. LetG be a nontrivial finite group. (1) IfK is a minimal normal subgroup ofG, andL is any normal subgroup ofG, then eitherK L or K,L =K L. 27 ≤ (2) There� exist� minimal× normal subgroupsK ,...,K ofG such that soc(G)=K K . 1 m 1 × · · · m (3) Every minimal normal subgroupK ofG is a direct productK=T 1 T k where theT i are simple normal subgroups ofK which are conjugate inG. × · · · (4) If the subgroupsK i in (2) are all nonabelian, thenK 1,...,Km are the only minimal normal sub- groups ofG. Proof. (1) #inceK L�G, the minimality ofK implies that eitherK L orK L= 1 . $n the latter case, since∩ K andL are both normal, we have that K,L =≤KL=K∩L. { } (2) %ecauseG is &nite we can &nd a setS= K ,...,K � of� minimal normal× subgroups which is { 1 m} ma'imal with respect to the property thatH (= S =K 1 K m. )e must show thatH contains all minimal normal subgroups ofG, and� so� is equal to× soc(· · · ×G). This follows immediately from (1). (3) +etT be a minimal normal subgroup ofK and observe that all conjugatesT g, forg G, are also minimal normal subgroups ofK. Choose a setS= T ,...,T of these conjugates∈ which is { 1 m} ma'imal with respect to the property thatL (= S =T 1 T m. Then, arguing $ala (2), we see thatL contains all of the conjugates ofT under� G� and so×L · ·G ·. × %ut, since 1 <L K and � { } ≤ K is minimal normal, we conclude thatK=L=T 1 T m. Note, &nally, that, forT i to be minimal normal inT T , we must haveT simple.× · · · × 1 × · · · × m i (4) +etK be a minimal normal subgroup ofG that is distinct fromK 1,...,Km, then, applying (1) withL=K , we &nd that K,L =K L and, in particular,K centralizes each of theK . Thus i � � × i K Z(soc(G)). However if eachK i is nonabelian, then (3) implies thatZ(K i) = 1 and we have a contradiction,≤ as required. { } � 0bserve that if a minimal normal subgroupK is abelian, thenK is an elementary-abelianp-group for some primep. 28 (E7.5)Suppose thatG is elementary abelian. "ow many minimal (resp. maximal) normal sub- groups doesG have? 26i.e. soc(G) is invariant under any automorphism ofG. This is because any automorphism ofG must permute the minimal normal subgroups ofG. 27I am considering the internal direct product here, a special case of the internal semidirect product that we have already seen. 28Recall that an elementary-abelianp-group is defined to be a group that is isomorphic toC C , wheren is finite, p × · · · × p n p is a prime, andC p is a cyclic group of orderp. � �� � FINITE PERMUTATION GROUPS AND FINITE CLASSICAL GROUPS 35 7.1. Finite groups with elementary abelian socle. In this sectionG is a &nite group with an elementary-abelian socle. )e write V (= soc(G)=C C . p × · · · × p d 0bserve thatV has a natural structure as a vector space� over�� F p�, the &eld of orderp. This allows us to ma2e the following assertion. Lemma 7.2. G/CG(V) is isomorphic to a subgroup of 3+(V), and G/C G(V) acts onV via multiplication (on the right) by matrices. Proof. Consider the action ofGonV by conjugation. +emma !. implies that this induces a homomorphism φ(G Aut(V ) where we viewV is an ob,ect from Group. 4urthermoreC G(V) is the 2ernel ofφ and, in particular,→ it is a normal subgroup ofG. Now, sinceC G(V ) is the 2ernel of the conjugation action, the &rst isomorphism theorem of groups implies that G/V is isomorphic to a subgroup of Aut(V ). Now the result follows from (E7.6) below.� (E7.6) $etV be an elementary#abelianp#group. (1) $etG be a group of automorphisms ofV as an object from Group. &rove thatG acts linearly onV , i.e. prove thatG is a group of automorphisms ofV as an object from Vect Fp . (2) $etG be a group of automorphisms ofV as an object from Vect Fp . Prove thatG is a group of automorphisms ofV as an object from Group. (3) Conclude that Aut(V ) # $L(V), whether we consider it an object of Group or of Vect Fp . +et us strengthen our supposition( let us suppose thatG splits overV , i.e. that there e'ists a subgroup H < G, such thatG=V�H. To state the structure result in this case, we need a definition. Given a vector spaceV over a &eldK, define A3+(V)(= (g,v) v K d, g 3+(V) =K d � 3+(V). { | ∈ ∈ } Here we writeK d to mean the additive group whose elements ared-tuples with entries fromK. Multipli1 cation is defined in the usual way for a semidirect product( g2 (g1, v1)(g2, v2) (= (g1, g2, v1 v2) n g where, forv K andg 3+ d(K), we definev to be the product ofv (thought of as a row vector) with g, a matri'.∈ ∈ Note that, just as with 3+(V ), we will write A3+d(K) as a pseudonym for A3+(V ). 4urthermore, if K=F p is &nite, then we will write A3+d(p) as a synonym for A3+d(Fp). Proposition 7.3. Suppose thatG is a finite group with socleV of orderp d for some primep. IfG splits overV , thenG is isomorphic to a subgroup of A3+ d(p). Proof. +etH be a subgroup ofG such thatG=V�H. Consider the action ofH onV by con,ugation. )e claim thatC H (V ) is a normal subgroup ofG8 indeed it is enough to prove that it is a normal subgroup ofH, sinceG=VH andC H (V ) is certainly normalized byV. To prove the claim, ta2eh C H (V),h 1 H andv V . Now observe that ∈ ∈ h ∈ (h 1 ) h1 1 h1 v = (h )− v(h ) 1 1 1 = (h1− hh1)− v(h1− hh1) 1 1 1 =h 1− (h− (h1vh1− )h)h1 1 1 =h 1− (h1vh1− )h1 =v. h1 Thush C H (V ) and the claim is proved. But now, sinceH V= 1 , we know thatC H (V) V= 1 and so, since∈ V is the socle, we conclude thatC (V)= 1 . ∩ { } ∩ { } H { } Now considerC G(V ). Ifg G, theng=vh for a uni*uev V andh H andg C G(V) if and only ifh C (V ). In particular,∈ ∈ ∈ ∈ ∈ G CG(V)= V.C H (V)=V. 36 NICK GILL Now +emma 7. implies that G/V is isomorphic to a subgroup of 3+(V ). Moreover the groupH ∼= G/V acts onV by right multiplication by matrices; in other wordsV�H is isomorphic to a subgroup of A3+ d(p) as required. � (E7.7)Suppose thatK is finite. Prove that soc("$% (K)) # Kd and, moreover, that soc("$% (K)) is a minimal normal • d ∼ d subgroup of "$%d(K). Suppose that soc("$%d(K) G "$% d(K). (nder what conditions is soc("$%d(K))a • minimal normal subgroup of≤G?≤ (E7.8)Describe the structure of "$%1(p) for a primep. 7.2. The socle of a primitive permutation group. Theorem 7.4. IfG is a finite primitive subgroup of #ym(n), andK is a minimal normal subgroup ofG, then exactly one of the following holds: (1) for some primep and some integerd,K is a regular elementary abelian group of orderp d, and soc(G) =K=C G(K). (2)K is a regular non-abelian group,C G(K) is a minimal normal subgroup ofG which is permutation isomorphic toK, and soc(G) =K C G(K). (3)K is non-abelian and soc(G)=K.× Proof. +etC=C G(K). #inceC�G, eitherC=1orC is transitive. #inceK is transitive, Lemma !.7 implies thatC is semiregular, and hence eitherC=1orC is regular. #uppose thatC = 1. ClearlyK is non-abelian. 4urthermore Theorem 7.1 (1) implies thatK is the only minimal normal subgroup ofG and so soc(G)=K and conclusion (3) of the result holds. #uppose instead thatC is regular. Then (5!. !) implies thatC (C) is regular. #inceK Sym(Ω) ≤ CSym(Ω)(C) andK is transitive, we conclude thatK=C Sym(Ω)(C). Similarly,C=C Sym(Ω)(K) and, by (5!. "),C andK are permutation isomorphic. Now, sinceC is regular, we conclude thatC is a minimal normal subgroup ofG (since any proper subgroup ofC is intransitive). Now Theorem 7.1 (1) states that every minimal normal subgroup ofG distinct fromK is contained inC. Thus soc(G) =KC which e*ualsK orK C depending on whether C K or not. × ≤IfC K, thenC=K,K is abelian, soc(G)=K and conclusion (1) holds. IfC K, then conclusion ≤ �≤ (2) holds. � (E7.9) Suppose thatG is a maximal primitive subgroup of &ym(n).