The largest conjugacy class size and the nilpotent subgroups of finite groups

Theorem 1.1.

Let H be a nilpotent subgroup of a nonabelian finite group G, and let bcl⁡(G) be the largest conjugacy class size of the group G. Let π=π⁢(|H|) be the set of prime divisors of |H|. Then |H⁢Oπ⁢(G)/Oπ⁢(G)|<bcl⁡(G).

Lemma 2.1.

Let q=rf, where r is a prime and f is a positive integer. Then:

1. q>2⁢d⁢f unless q<7, where d=(2,q-1),

2. rf+1>4⁢f unless rf∈{21,22,23,31},

3. rf-1>2⁢f unless rf∈{21,22,31}.

Proof.

This elementary calculation is omitted. ∎

Lemma 2.2.

Let G be a nilpotent permutation group of degree n. Then |G|≤2n-1.

Proof.

This result is well known, see for example [2]. ∎

Lemma 2.3 ([9, Lemma 3]).

Let G be a finite solvable group such that G/F⁢(G) is nilpotent. Then k⁢(G)≤|F⁢(G)|.

Lemma 2.4.

Let a finite nilpotent group G act faithfully and coprimely on an abelian group V. Then the largest conjugacy class size in the semidirect product G⋉V is greater than |G|.

Proof.

By Lemma 2.3, we know that the number of conjugacy classes in G⋉V is less than or equal to |V|. Also the identity is a conjugacy class of size 1, and the result follows. ∎

Proposition 2.5.

Let G be one of the nonabelian simple groups and H a nilpotent subgroup of Aut⁡(G). Then bcl⁡(G)>2⁢|H|.

Proof.

This will be proved by combining Lemmas 2.6, 2.7 and 2.9. ∎

Lemma 2.6.

Let G be one of the alternating groups Altn,n≥5, and H a nilpotent subgroup of Aut⁡(G). Then bcl⁡(G)>2⁢|H|.

Proof.

For n=5, we have Aut⁡(Alt5)≅Sym5, |H|≤8, bcl⁡(Alt5)=20, and the result is clear. For n=6, we have Aut⁡(Alt6)≅K, where |K/Alt6|=4, and so |H|≤25. Let α∈Alt6 be of order 5. Then bcl⁡(Alt6)≥|αAlt6|=72, and the result follows.

For the case when n≥7, set

Then α∈Altn, and

Note that Aut⁡(Altn) is isomorphic to Symn, and that |H|≤2n-1 by Lemma 2.2. Since 12⁢(n-1)!>2n for n≥7, we get the required result. ∎

Lemma 2.7.

Let G be one of the sporadic groups or Tits group, and H a nilpotent subgroup of Aut⁡(G). Then bcl⁡(G)>2⁢|H|.

Proof.

This follows by [1] and [20], see also [21, Table 1]. ∎

Lemma 2.8 ([20, Table 3]).

Let G be a nonabelian simple group of Lie type over a field of characteristic r. Then |G|nil=|G|r unless one of the following holds:

1. G≅A1⁢(2n), |G|nil=2n+1,

2. G≅A1⁢(q), q=-1+2n, |G|nil=|G|2=2n,

3. G≅A22⁢(3), |G|nil=|G|2=32,

4. G≅A32⁢(2)≅B2⁢(3), |G|nil=|G|3=81.

Lemma 2.9.

Let G be one of the nonabelian simple groups of Lie type and let H be a nilpotent subgroup of Aut⁡(G). Then bcl⁡(G)>2⁢|H|.

Proof.

With the notation as above, we know by [7, Theorem 2.2.7] that Or′⁢(G~σ) is a nonabelian simple group with the following exceptions:

The first four of these exceptions are solvable. If G≅F42⁢(2)′, then the required result follows by Lemma 2.7. We regard B2⁢(2)′ as A1⁢(9), G2⁢(2)′ as A22⁢(3), and G22⁢(3)′ as A1⁢(8). Now we may assume that G=Or′⁢(G~σ)=L⁢(q).

Set G1=G~σ. By [3, Theorem 3.1], G1 has a maximal torus T such that T∩G is a self-centralizing cyclic subgroup of G. Note that the order of T is known, see for example [13, Table 1]. Write T0=T∩G=〈a〉. Then |aG|=|G|/|T0|. To see that bcl⁡(G)>2⁢|H|, it suffices to show that

We verify the result by going through the classifications.

Case 1. Assume that G=An⁢(q), q=rf. We consider two subcases.

(1) Suppose that n=1. Then q=rf≥4, |G|=q⁢(q+1)⁢(q-1)⁢d-1, where d=(2,q-1), and |Out(G)|=df. Since A1⁢(4)≅A1⁢(5)≅Alt5 is treated in Lemma 2.6, we may assume that q≥7. Then q>2⁢d⁢f by Lemma 2.1. Note that G has a self-centralizing cyclic subgroup T0 of order (q-1)⁢d-1, and that |G|nil≤q+1 by Lemma 2.8. We get that

and we are done.

(2) Suppose that n≥2. In that case, we have |G|=d-1⁢qn⁢(n+1)/2⁢m, where m=∏i=1n(qi+1-1), d=(n+1,q-1), and |Out(G)|=2fd. Note that

by [13, Table 1] and Lemma 2.8. To see that |G|>2|T||Out(G)||G|nil, that is, ∏i=1n(qi-1)>4⁢d2⁢f, it suffices to show that (q2-1)⁢(q-1)>4⁢d2⁢f when n=2. If q∉{2,22,23,3}, then rf+1>4⁢f by Lemma 2.1, and (q2-1)⁢(q-1)>4⁢d2⁢f as wanted. If q∈{2,22,23,3}, then the required result follows by [1].

Case 2. Assume that G≅An2⁢(q) where n≥2. Note that if n=2, then q>2. Set m=∏i=1n(qi+1-(-1)i+1), q2=rf, and d=(n+1,q+1). Then we have |G|=d-1⁢m⁢qn⁢(n+1)/2 and |Out(G)|=df. If G is isomorphic to A22⁢(3) or A32⁢(2), then the required result follows by [1]. Now let us consider the remaining cases, we have by Lemma 2.8 that |G|nil=|G|r. Note that

by [13, Table 1]. We need only to show that

that is, ∏i=1n(qi-(-1)i)>2⁢d2⁢f. To do this, it suffices to show that q-1>2⁢f when n=2. We know that rf-1>2⁢f unless rf=22,3, and for these exceptional cases, the result can be verified by direct calculations.

Case 3. Let G be of type Bn⁢(q), where n≥2 and q=rf. Set

Then |G|=d-1⁢qn2⁢m and |Out(G)|=dfg, where d=(2,q-1), and g≤2. Note that |T|=qn+1 by [13, Table 1]. Thus it suffices to show that (q2-1)2>2⁢d2⁢f. Using Lemma 2.1 (3), one can check this by direct calculations.

Case 4. Let G be of type Cn⁢(q), where n≥3. This proof is almost identical to Case 3 and so is omitted.

Case 5. Let G be of type Dn⁢(q), where n≥4 and q=rf. Set

Then

Note that |T|=qn-1 by [13, Table 1]. It suffices to show that

One checks directly that the inequality holds.

Case 6. Let G be of type Dn2⁢(q2), where n≥4. Set

Then |G|=d-1⁢m⁢qn⁢(n-1)⁢(qn+1) and |Out(G)|=d⋅f. Note that |T|=qn+1 by [13, Table 1]. It suffices to show that

Again this is easy to check.

Case 7. For the remaining simple groups of Lie type, one can check the result by examining [13, Table 1]. ∎

Theorem 1.1.

Let H be a nilpotent subgroup of a finite nonabelian group G, and π=π⁢(|H|) the set of prime divisors of |H|. Then |H⁢Oπ⁢(G)/Oπ⁢(G)|<bcl⁡(G).

Proof.

If H=1, then π is an empty set and Oπ⁢(G)=1. Then the result is clear since G is nonabelian and thus we may assume H>1. Suppose that Oπ⁢(G)≠1. If G/Oπ⁢(G) is abelian, then H≤Oπ⁢(G) and |H⁢Oπ⁢(G)/Oπ⁢(G)|=1. Thus we have bcl⁡(G)>|H⁢Oπ⁢(G)/Oπ⁢(G)|, and we are done. If G/Oπ⁢(G) is nonabelian, then induction yields the required results. Thus we may assume that Oπ⁢(G)=1.

Let F*⁢(G) be the generalized Fitting subgroup of G. Then we know that F*⁢(G)=F⁢(G)⁢E⁢(G). Let Z=Z⁢(E⁢(G)). By the proof of [6, Lemma 3.15] and by [6, Lemma 3.16] we know that F⁢(G/Z)=F⁢(G)/Z, E⁢(G/Z)=E⁢(G)/Z, F*⁢(G/Z)=F*⁢(G)/Z, and CG/Z⁢(E⁢(G/Z))=CG⁢(E⁢(G))/Z. As Oπ⁢(G)=1, we know that Z is a π′-group.

We denote K=G/Z, H¯=H⁢Z/Z, and we know that H¯>1. We will show that bcl⁡(K)>|H¯|=|H|. Since Oπ⁢(G)=1, we know that F⁢(K) is a π′-group.

We first assume that E⁢(K)=1. In this case, we have 𝐂H¯⁢(F⁢(K))≤F⁢(K) and thus 𝐂H¯⁢(F⁢(K))=1. We have 𝐂H¯⁢(F⁢(K)/Φ⁢(F⁢(K)))=1. We consider the semidirect product H¯⋉(F⁢(K)/Φ⁢(F⁢(K))). Since H¯>1, it follows by Lemma 2.4 that there exists some x¯∈L=H¯⁢(F⁢(K)/Φ⁢(F⁢(K))) such that bcl⁡(L)≥|H¯|. Thus, since L is a quotient group of H¯⁢F⁢(K), we know that bcl⁡(G)≥bcl⁡(L)≥|H¯|, and the theorem holds in this case.

We now assume that E⁢(K)>1. Clearly E⁢(K) is a direct product of nonabelian simple groups. We write E⁢(K)=E1×E2×⋯×Em, where Ei is a minimal normal subgroup of K and is also a direct product of ki isomorphic nonabelian simple groups Li.

We denote Nm=𝐂K⁢(Em), Km=K/Nm, and we define Hm=H¯⁢Nm/Nm. Since Em≅(Em×Nm)/Nm, we can consider Em as a subgroup of Km, and we know that Km is embedded in Aut⁡(Em)≅Aut⁡(Lm)≀Symkm. Recursively, we define Ni=𝐂Ni+1⁢(Ei), Ki=Ni+1/Ni, and Hi=(H¯∩Ni+1)⁢Ni/Ni for 1≤i≤m-1. Since Ei≅(Ei×Ni)/Ni, we can consider Ei as a subgroup of Ki, and we know that Ki is embedded in Aut⁡(Ei)≅Aut⁡(Li)≀Symki. So we have a chain of quotients, and we know that N1=𝐂N2⁢(E1)=𝐂K⁢(E⁢(K)) and we define H0=H¯∩N1. Thus |H|=|Hm|⋅|Hm-1|⁢⋯⁢|H1|⋅|H0|.

Clearly F⁢(K)≤N1 and N1∩E⁢(K)=1. By [6, Lemma 3.13 (i)],

We have 𝐂H0⁢(F⁢(K))≤F⁢(K) and thus 𝐂H0⁢(F⁢(K))=1. Since the action of H0 on F⁢(K) is coprime, 𝐂H0⁢(F⁢(K)/Φ⁢(F⁢(K)))=1.

If H0=1, then clearly bcl⁡(N1)≥|H0|. If H0≠1, it follows by Lemma 2.4 that there exists some x¯∈Q=H0⁢(F⁢(K)/Φ⁢(F⁢(K))) such that bcl⁡(Q)>|H0|. Therefore, since Q is a quotient group of H0⁢F⁢(K), we know that

and so that bcl⁡(N1)>|H¯0|.

We know that Ki is embedded in Aut⁡(Ei)=Aut⁡(Li)≀Symki. For 1≤i≤m, set Ji:=Ki∩Aut(Li)ki. Note that Ei can be viewed as a subgroup of Ki and therefore Ei is a subgroup of Ji, and we denote that

Also, Ki/Ji is a permutation group on ki letters.

Let vi⁢j∈Li⁢j such that vi⁢jLi⁢j=bcl⁡(Li) and set vi=vi⁢1⁢⋯⁢vi⁢ki. Clearly vi∈Ei and

Thus,

by Lemma 2.2 and Proposition 2.5. Let v=v1⁢⋯⁢vm and we have that

Note that v⋅x is an element of E⁢(K)×N1. It follows that

and we are done. ∎

Theorem 3.1.

Let P be a Sylow p-subgroup of a finite nonabelian group G. Then |P/Op⁢(G)|<bcl⁡(G).

Remark 3.2.

We provide a family of examples to show that our result is the best possible. Let G=K⋊V, where |V| is a Fermat prime, |K|=|V|-1=2n and K acts fixed point freely on V. Note that bcl⁡(G)=|V| and

Remark 3.3.

We note that by using [9, Theorem 10], one may obtain directly that for an arbitrary finite group, |G:𝐅(G)|≤bcl2(G).

Remark 3.4.

We observe that the proofs in [11, 13, 21, 17] have been greatly influenced by the character degree analogues (see for example [15, 16]). The main idea in all those papers is that, after the reduction steps, one tries to use the orbit theorems [10, 12] to estimate the largest conjugacy class size or the largest character degrees. Unfortunately, the largest-orbit theorems of this type have their limits, and that is the main reason why the results in [11, 13, 21] are not ideal. However, it turns out that after the reduction steps, the remaining structure is suitable in applying the k⁢(G⁢V)-type results, namely one has some knowledge on the average of all the conjugacy class sizes. It is a little surprising to see that, in this particular problem, the estimate on “average” is better than the estimate on the “maximum”. Our understanding on this phenomenon is that, by using the orbit theorem, one tries to find a large conjugacy class that lives in the Fitting subgroup but mostly likely the conjugacy class of the largest size could well live outside the Fitting subgroup. Another observation is that the k⁢(G⁢V) bound is very strong, even when the acting group is nilpotent. On the other hand, it seems that for character degrees, the orbit theorem approach works pretty well.