Bounding the number of classes of a finite group in terms of a prime

Theorem 1.1.

There exists a constant c>0 such that, for any finite group G whose order is divisible by the square of a prime p, we have k⁢(G)≥c⁢p.

Lemma 2.1.

Let H be a finite group and V a finite, faithful, completely reducible H-module over a finite field of characteristic p. Assume that H has no composition factor isomorphic to an alternating group of degree larger than (log⁡p)3 and has no composition factor isomorphic to a simple group of Lie type defined over a field of characteristic p. Put pn=|V|. Then H has an abelian subgroup of index at most (c1⁢log⁡p)7⁢(n-1) for some universal constant c1>1.

Proof of Lemma 2.1.

Assume first that V is a primitive and irreducible H-module. We use the following structure result which is implicit in the proofs of [6] (see for example the proof of [6, Theorem 9.1]). Let F be the largest field such that H embeds in Γ⁢LF⁢(V). Let C be the subgroup of non-zero elements in F. We claim that |H/(H∩C)|≤(c1⁢log⁡p)7⁢(n-1) for some universal constant c1>1. For this, we may assume that C≤H.

Let H0 be the centralizer of C in H, and let R be a normal subgroup of H contained in H0 minimal with respect to not being contained in C (if such exists). There are two possibilities for R. It is of symplectic type and |R/Z⁢(R)|=r2⁢a for some prime r and integer a such that r divides |F|-1 or R is a central product of t isomorphic quasisimple groups.

Choose a maximal collection J1,…,Jm of such non-cyclic normal subgroups in H0 which pairwise commute (if such exist). Let J be the central product of the subgroups J1,…,Jm. Then H0/(C⋅Sol⁢(J)) embeds in the direct product of the automorphism groups of Ji/Z⁢(Ji), where Sol⁢(J) denotes the solvable radical of J. (Note that, in the proof of [6, Theorem 9.1], it was falsely asserted that H0/C embeds in the direct product of the automorphism groups; however, this did not affect the proof of [6, Theorem 9.1] nor [6, Theorem 10.1].)

Let W be an irreducible constituent of V for the normal subgroup J of H (provided that J is non-trivial). Since H is primitive on V, it follows that J acts homogeneously on V by Clifford’s theorem. Let E=EndF⁢J⁢(W). Now we have W≅U1⊗⋯⊗Um, where Ui is an absolutely irreducible E⁢Ji-module by [13, Lemma 5.5.5]. Notice that E may be viewed as a subfield of EndF⁢J⁢(V), and since J is normal in H, the multiplicative group of E is normalized by H. Our choice of F implies that E=F. If Ji is of symplectic type with Ji/Z⁢(Ji) of order ri2⁢ai, then dim⁡Ui=riai. If Ji is a central product of t isomorphic quasisimple groups Qi,j with 1≤j≤t, then Ui≅Ui,1⊗⋯⊗Ui,t, where Ui,j is an absolutely irreducible (faithful) F⁢Qi,j-module for every j with 1≤j≤t, by [13, Lemma 5.5.5 and Lemma 2.10.1].

Let |F|=pf and d=dimF⁡V. The product of the orders of all abelian composition factors in any composition series of the factor group H/C is less than f⋅d2⁢log⁡d+3≤n2⁢log⁡n+4 by [6, Theorem 10.1] and its proof. This is at most (c2⁢log⁡p)n-1 for some constant c2>2. We may now assume that J≠1 and n>1.

Let b⁢(X) denote the product of the orders of all non-abelian composition factors in any composition series of a finite group X. Since |H/C|≤(c2⁢log⁡p)n-1⁢b⁢(H), we proceed to bound b⁢(H).

Without loss of generality, assume that J1,…,Jk are groups of symplectic type with k≥0 and |Ji/Z⁢(Ji)|=ri2⁢ai for some primes ri and integers ai, and assume that Jk+1,…,Jm are groups not of symplectic type. For each ℓ with k+1≤ℓ≤m, let Jℓ be a central product of tℓ copies, say Qℓ,1,…,Qℓ,tℓ, of a quasisimple group Qℓ. In this case, Uℓ≅Uℓ,1⊗⋯⊗Uℓ,tℓ, where Uℓ,j is an irreducible (faithful) Qℓ,j-module for every j with 1≤j≤tℓ. Using this notation, we may write the following:

Since H/H0 and C⋅Sol⁢(J) are solvable, b⁢(H)=b⁢(H0/(C⋅Sol⁢(J))). Recall from the third paragraph of this proof that the group H0/(C⋅Sol⁢(J)) embeds in the direct product of the automorphism groups of the Ji/Z⁢(Ji). There exists a chain of subnormal subgroups

such that Ni-1/Ni≤Aut⁢(Ji/Z⁢(Ji)) for every i with 1≤i≤m. These give

Since ∏i=1kriai≤n by (2.1), we have

We see by Schreier’s conjecture that, for every ℓ with k+1≤ℓ≤m, we have b⁢(Nℓ-1/Nℓ)≤|Tℓ|⋅b⁢(Qℓ)tℓ, where Tℓ is some permutation group of degree tℓ having no composition factor isomorphic to an alternating group of degree larger than (log⁡p)3. Now |Tℓ|≤(2⁢log⁡p)3⁢(tℓ-1) by [18, Corollary 1.5]. Using the fact that ∑ℓ=k+1mtℓ≤log⁡n (see (2.1)), we have

It follows by (2.2), (2.3) and (2.4) that

for some constant c3>1.

Let T be a quasisimple group with T/Z⁢(T) not isomorphic to an alternating group of degree larger than (log⁡p)3 and not isomorphic to a simple group of Lie type defined over a field of characteristic p. Let U be any finite, faithful FT-module over the finite field F of order pf. Put |F|s=|U|. We claim that

for some universal constant c4>1. We use [15]. A consequence of [13, (5.3.2), Corollary 5.3.3 and Theorem 5.3.9] is that if T/Z⁢(T) is a simple group of Lie type in characteristic different from p, then |T/Z⁢(T)|<(c4⁢log⁡p)3⁢(s-1) for some constant c4>1. By choosing c4 to be at least the maximum of the size of the Monster and the largest value of r! for which r!≥rr-5, where r is a positive integer, our bound on |T/Z⁢(T)| extends to the case when T/Z⁢(T) is a sporadic simple group or T/Z⁢(T) is an alternating group of degree r with r!≥rr-5. If T/Z⁢(T) is an alternating group of degree r≤(log⁡p)3 such that r!<rr-5, then

where the last inequality follows from [13, (5.3.2), Corollary 5.3.3 and Proposition 5.3.7]. This proves our claim.

For every ℓ with k+1≤ℓ≤m, define sℓ≥2 by

Using (2.6) and (2.1), we find that

We have b⁢(H)<(c3⁢c4⁢log⁡p)6⁢(n-1) by (2.5) and (2.7). Thus

Finally, set c1=c2⁢c3⁢c4>2.

This finishes the proof of the lemma in case V is a primitive and irreducible H-module.

Let H be a counterexample to the statement of the lemma with dim⁡V minimal and with c1 as before. Put f⁢(p)=(c1⁢log⁡p)7.

We claim that V must be an irreducible H-module. Assume that V=V1⊕V2, where V1 and V2 are non-trivial (completely reducible) H-modules. Let H1 be the action of H on V1, and let H2 be the action of H on V2. The groups H1 and H2 are factor groups of H and thus have no non-abelian composition factor which is not a composition factor of H. The group H may be viewed as a subgroup of H1×H2. Since H is a counterexample with dim⁡V minimal, there exist an abelian subgroup A1 in H1 of index at most f⁢(p)m-1 and an abelian subgroup A2 in H2 of index at most f⁢(p)n-m-1, where pm=|V1|. The group A=(A1×A2)∩H is an abelian subgroup of H. Moreover,

This is a contradiction. Thus V is an irreducible H-module.

We claim that V cannot be an imprimitive H-module. Let V=V1+⋯+Vt with t>1 be an imprimitivity decomposition for V with each Vi a subspace in V, and let N be the normal subgroup of H consisting of all elements leaving every Vi invariant. The group N acts completely reducibly on V and thus also on each Vi by Clifford’s theorem. For every i with 1≤i≤t, let Hi be the action of N on Vi. The group H/N may be viewed as a permutation group of degree t. In particular, H may be viewed as a subgroup of a full wreath product of the form W=(H1×⋯×Ht):Sym⁢(t). Since H is a counterexample with dim⁡V minimal, there exists an abelian subgroup Ai in Hi, for every i with 1≤i≤t, such that |Hi:Ai|≤f(p)(n/t)-1. The group A1×⋯×At is contained in W. Thus A=(A1×⋯×At)∩N is an abelian subgroup in H. As before,

The permutation group H/N of degree t has no composition factor isomorphic to an alternating group of degree larger than (log⁡p)3. It follows that

by [18, Corollary 1.5]. We thus have |H:A|<f(p)n-tf(p)t-1=f(p)n-1 by (2.8) and (2.9). A contradiction.

This finishes the proof of the lemma. ∎

Proposition 2.2.

There exists a universal constant c5>0 such that if G is a finite group having an elementary abelian minimal normal subgroup V of p-rank at least 2 and |G/V| is not divisible by p2, then k⁢(G)≥c5⁢p.

Proof.

Since k⁢(G)≥k⁢(G/V)+n⁢(G,V)-1 by Clifford’s theorem, it is sufficient to show that k⁢(G/V)+n⁢(G,V)≥c6⁢p for some universal constant c6>0. For this latter claim, we may assume that G/V acts faithfully on V, that is, V is a faithful and irreducible H:=G/V-module. This is because

We may assume that p is sufficiently large.

Every non-abelian (simple) composition factor of H (provided that it exists) has order coprime to p, except possibly one which has order divisible by p (but not by p2). There are the following possibilities for a non-abelian composition factor S of H: (i) S is an alternating group; (ii) S is a simple group of Lie type in characteristic different from p; (iii) S≅PSL⁢(2,p); (iv) S is a sporadic simple group.

Suppose that such a composition factor S exists. We have k⁢(H)≥k*⁢(S) by [21, Lemma 2.5]. Since k*⁢(PSL⁢(2,p))≥(p-1)/4, by considering diagonal matrices in SL⁢(2,p), we may exclude case (iii) by choosing c5<15 (since we are assuming that p is sufficiently large). Let S be an alternating group of degree r≥5. Since |Out⁢(S)|≤4, we have k*⁢(S)≥k⁢(S)/4. Since S is a normal subgroup of index 2 in the symmetric group of degree r, we have k⁢(S)≥π⁢(r)/2 where π⁢(r) denotes the number of partitions of r. We thus find that k*⁢(S)≥c7r for some constant c7>1. If r>(log⁡p)3, then k*⁢(S)>p for sufficiently large p. Thus we assume that every alternating composition factor of H has degree at most (log⁡p)3.

The group H contains an abelian subgroup A with

by Lemma 2.1. Furthermore, k(H)≥k(A)/|H:A|=|A|/|H:A| by [4, p. 502] and n⁢(G,V)≥|V|/|H|. These give

Since the real function g⁢(x)=x+(|V|/x) takes its minimum in the interval [1,|V|] when x=|V|, we find that k⁢(H)+n⁢(G,V)>2⋅|V|(1/2)-o⁢(1)>p for sufficiently large p, unless |V|=p2.

Let |V|=p2. Note that, in [9, p. 661 and 662], it is shown that if G is solvable, we have

Thus we may assume that G is non-solvable. In this case, H/Z⁢(H) is either Alt⁢(5) or Sym⁢(5) (given that case (iii) above cannot occur) by [2, Section XII.260] or [10, Hauptsatz II.8.27]. Also, |Z⁢(H)|<p since H is non-solvable by assumption. Thus there exists a constant c8>0 such that k⁢(G)≥n⁢(G,V)≥|V|/|H|>c8⁢p.

This finishes the proof of the proposition. ∎

Lemma 3.1.

Let p,q∈N+∖{1} such that p∣qi+(-1)a and p∣qj+(-1)b for some i,j∈N+ and some a,b∈{0,1}. If (i,a)≠(j,b), then

Proof.

Without loss of generality, we may assume that j≤i. By our assumptions, since p≤qj+1, it is sufficient to show that p≤qi-j+1. We have

Assume first that i≠j and a=b. Then p∣qi-qj=qj(qi-j-1), and so p≤qi-j-1. If a≠b, then

and so p≤qi-j+1. ∎

Lemma 3.2.

Suppose that P⁢(x) is a polynomial admitting a factorization of the form P+⁢(x)⁢P-⁢(x), where

for some sets of integers S+ and S- and positive integers ki+ and ki-. Set

Assume that ki+=ki-=1 for every index i strictly larger than m/2. If p is an odd prime such that p2∣P(q) for some integer q≥2, then p≤qm/2+1.

Proof.

Let p be an odd prime such that p2∣P(q) for some positive integer q≥2. Let i∈S+∪S- be such that p∣qi+1 or p∣qi-1. If i≤m/2, the assertion is clear, so assume that i>m/2. If p2∣qi+1 or p2∣qi-1, then

Otherwise, there exists j∈S+∪S- distinct from i such that

By Lemma 3.1, p≤qmin⁡{i,j,|i-j|}+1. Observe that min⁡{i,j,|i-j|}≤m/2. For a proof of this observation, we may assume that i≤j≤m, and so i or j-i is at most m/2. ∎

Lemma 3.3.

Let S be a finite simple group of Lie type of Lie rank r defined over a field of size q as in Table 1. If p is an odd prime such that p∤q and p2∣|S|, then p≤q(r+1)/2+1 if r>8 and p≤qr-12+1 if r≤8.

Proof.

Observe that |S| divides P-⁢(q)⁢P+⁢(q) times a suitable power of q, so p2∣P-(q)P+(q). If m is as in the statement of Lemma 3.2, then p≤qm/2+1. According to Table 1, m≤r+1 if r>8 and m≤2⁢r-1 if r≤8. The result follows. ∎

Lemma 3.4.

Let S be a finite simple group of Lie type of Lie rank r defined over a field of size q as in Table 2. There exists a constant c10 such that if p is an odd prime with p∤q and p2∣|S|, then p≤c10⋅qr-12.

Proof.

Let P⁢(q) be as in Table 2. Notice that p2∣P(q). If p2 divides any of the three, four and eight factors in the factorizations of P⁢(q) in Table 2, in the respective three cases, then the statement holds. The statement also holds in case S=F42⁢(q) when p2∣(q-1)2. We may assume that there are two distinct factors P1⁢(q) and P2⁢(q) in the factorization of P⁢(q) given in Table 2 which are divisible by p. Hence p∣P1(q)-P2(q).

Let S be B22⁢(q), where q=22⁢t+1. In this case, |P1⁢(q)-P2⁢(q)|≤2⁢2⁢q. Similarly, if S is G22⁢(q), then p≤2⁢3⁢q.

Let S be F42⁢(q), where q=22⁢t+1. Assume first that P1⁢(q) and P2⁢(q) have the same degree. In this case, |P1⁢(q)-P2⁢(q)|≤2⁢2⁢q3. Otherwise, p divides a factor of degree 1, and so p≤q+2⁢q+1. In any case, the result follows. ∎

Lemma 3.5.

Let S and r be as in Table 3. Then both k⁢(D43⁢(q)) and k⁢(E62⁢(q)) are at least c11⋅qr+2 for some constant c11>0. We also have