Primitive characters of odd order groups
Claudio Marchi




C. Marchi]Claudio Marchi

School of Mathematics 
 University of Manchester 

Oxford Road 
  M13 9PL 
 Manchester 

United Kingdom 

claudio.marchi@manchester.ac.uk














 [2000]20C15, 20C99



  



 Let G be a finite group of odd order. We show that if χ is an irreducible primitive character of G then for all primes p dividing the order of G there is a conjugacy class such that the p-part of χ(1) divides the size of that conjugacy class. We also show that for some classes of groups the entire degree of an irreducible primitive character χ divides the size of a conjugacy class.


Primitive characters of odd order groups
    [
    December 30, 2023
========================================





§ INTRODUCTION

Irreducible characters and conjugacy classes are related by many arithmetical properties. The most well-known is an immediate consequence of the Ito-Michler theorem saying that, if a prime number p divides the degree of an irreducible character then p divides the size of at least a conjugacy class. This result has been extended by Casolo and Dolfi in <cit.>, to show that if pq divides the degree of an irreducible character then there exists a conjugacy class whose size is divisible by pq.

In <cit.> Isaacs, Keller, Meierfrankenfeld and Moreto' conjectured the following:
[<cit.>,Conjecture C]
Let χ be an irreducible primitive character of a finite group G. Then χ(1) divides | cl_G(g)| for some element g∈ G.

In <cit.> Cassell proved that the conjecture holds for all finite simple groups and symmetric groups. One way of approaching the conjecture is to consider only the p-parts of the character degrees and in <cit.> the following was proved:
[<cit.>,Corollary D]
Let χ be an irreducible primitive character of a solvable group G and let p be a prime divisor of | G|. Then χ(1)_p divides (| cl_G(g)|_p)^3 for some element g∈ G.

By restricting our attention to odd order groups we are able to prove a result closer to the original conjecture:

Let χ be an irreducible primitive character of an odd order group G, p a prime dividing | G|. Then there is a g∈ G such that χ(1)_p divides | cl_G(g)|_p.

In particular this result gives a positive answer to the conjecture for primitive irreducible characters of prime power order of odd order groups.

We follow the same approach as in <cit.> but focusing on odd order groups we can use the symplectic structure of some modules to get sharper bounds on the order of Sylow subgroups. A crucial result in proving the theorem above is in fact the following:

Let V be a symplectic, faithful 𝔽_3G-module that is a direct sum of irreducible anisotropic modules and dim(V)=2n. If | G| is odd, O_3(G)=1 and P∈ Syl_3(G) we have j=log_3(| P|)≤β_3(2/3n).

We remark that it is not clear how to use the result on p-parts for proving the conjecture on the entire degree of a primitive character, since the conjugacy class whose size is divided by the p-part depends on the chosen prime. However we managed to prove that the conjecture holds also for the following family of groups:

Let G be a meta-nilpotent group, | G| odd, π(G)={ p_1,p_2}, χ∈Irr(G) primitive character. Then there is g∈ G such that χ(1) divides | cl_G(g)|.




§ SYMPLECTIC MODULES

In this section we collect all the basic information on symplectic modules and we prove theorems that will be crucial for proving the main result. From now on G will be a finite group, p will be a prime number, 𝔽 will be a field of positive characteristic, while 𝔽_p will be the field with p elements. 

A vector space V is symplectic if it carries an alternating bilinear form such that V^⊥=0.

An 𝔽G-module V is a symplectic module if it a symplectic space such that the symplectic form is G-invariant, i.e. ⟨ x^g,y^g⟩ =⟨ x,y⟩ for all x,y∈ V, g∈ G.


A submodule U of a symplectic 𝔽G-module V is totally isotropic if U≤ U^⊥. 

V is hyperbolic if it has a totally isotropic submodule with dimension 1/2dim(V).

V is anisotropic if it does not have non trivial totally isotropic submodules.



Let β_3,λ_3 be the following functions:

    β_3(x)=∑_i=0^∞⌊x/2· 3^i⌋    λ_3(x)=∑_i=1^∞⌊x/3^i⌋

If a,b,m,d,n are positive integers and η(x) is one of the previously defined functions, then the following hold 


  * η_3(a+b)≥η_3(a)+η_3(b)

  * η_3(md)≥ dη_3(m)+λ_3(d)

  * λ_3(n)≤ (n-1)/2

  * β_3(n)≤3/4n-1/2

  * β_3(2/3md)≥ dβ_3( 2/3m)+λ_3(d)


The following theorem follows the same proof of <cit.>.

Let V be a self dual,irreducible, faithful 𝔽G-module and | G| an odd order group. Let N be a normal subgroup of G such that

    V_N=e(V_1⊕...⊕ V_t)

with V_i irreducible 𝔽N-modules and t> 1. Then e is odd.


Take 𝔼 splitting field for G and N, and write the following:

    V^𝔼=⊕_j=1^sW^j
    
    V_i^𝔼=⊕_j=1^d_iV_i^j.

We remark that the V_i^js are distinct, and the W^j are distinct as well.

Let 𝔼_V be the subfield of 𝔼 obtaining adjoining to 𝔽 the values of the characters corresponding to the W_js. Actually, we just need to consider the values of the character corresponding to W_1 since the W_js are Galois conjugates.

We define 𝔼_V_i analogously. Again since the V_i are G-conjugates, then the irreducible constituents of V_k^𝔼 are conjugate of V_1^𝔼, so the fields 𝔼_V_i are all the same field. Thus, since d_i=|𝔼_V_i:𝔽|, we have d_i=:d for all i.

Now from <cit.> we have W^i_𝔽≃|𝔼:𝔼_V| V and (V_i^j)_𝔽≃|𝔼:𝔼_V_1| V_i. Suppose then that W^1,...,W^c are the constituents of V^𝔼 that have V_1^1 as a constituent after taking the restriction to N. Then

    W^1_N=...=W^c_N=e'(V_1^1⊕...⊕ V^k_t'_l_t'),

where the G-conjugate of V_1^1 are V_l_r^k_r, where l_1=1=k_1. But since we are talking of absolutely irreducible modules e' divides | G|, and in particular t' divides | G|. If now we take the restriction to 𝔽 in the previous equality we get (W^1_𝔽)_N=e'((V_1^1)_𝔽⊕...⊕ (V^k_t'_l_t')_𝔽). But then |𝔼:𝔼_V| V_N=e'|𝔼:𝔼_V_1|(V_1⊕ V_l_2...⊕ V_l_t'). Thus 

    et|𝔼:𝔼_V|=e't'|𝔼:𝔼_V_1|.

Take 𝔻 the subfield of 𝔼 generated by 𝔼_V and 𝔼_V_1. So dividing by |𝔼:𝔻| we have et|𝔻:E_V|=e't'|𝔻:𝔼_V_1|.

Suppose by contradiction that e is even, then |𝔻:𝔼_V_1| must be even as well. Let Γ=Gal(𝔻|𝔽). Γ is cyclic, so it contains just an involution π. Write 𝔼_V^*=Fix_Γ(𝔼_V) e 𝔼_V_1^*=Fix_Γ(𝔼_V_1).

Now |𝔻:𝔼_V_1|=|𝔼_V_1^*:1| is even and so 𝔼_V_1^* contains π, i.e. π fixes point-wise 𝔼_V_i, and so it fixes all the V_i^j. But π acts non trivially on 𝔻 and so on 𝔼_V. Thus π as a 𝔽 automorphism of 𝔼_V is the unique involution of Gal(𝔼_V|𝔽). Now V is self dual, so V^𝔼 is self dual as well. If W^1 were self dual we take ψ the corresponding Brauer character and by <cit.> we can find an irreducible ordinary character χ such that d_χψ is odd. But now G is an odd order group, so there are no non trivial self dual irreducible characters, thus χ=1_G and ψ is the trivial Brauer character. So W^1 is trivial and there is just one W^i, i.e. V^𝔼 is trivial and V is trivial, but this can not happen. So all the W^js are not self dual and π must send every W^i to its dual. 
 
Now since π fixes the V_i^j, applying π to the equation (<ref>) yields

    e'(V_1^1⊕...⊕ V_l_t'^k_t')   ≃ e'(π(V_1^1)⊕...⊕π(V_l_t'^k_t'))≃π(W_N^1)
       ≃Ŵ^1_N≃ e'(V̂_1^1⊕...⊕V̂_l_t'^k_t')

If V_1^1 is self dual then as before V_i is trivial and we have a contradiction. Thus no V_l_r^k_r is self dual.

We can then match each V_l_r^k_r with its dual and then t would be even, but t must divide | G| and we have the final contradiction.


If V is a symplectic, homogeneous, completely reducible module, V=eW, e> 1, with all irreducible submodules totally isotropic and self dual, then V admits a decomposition as orthogonal direct sum of symplectic hyperbolic modules, with dimension 2dim(W)

    V=H_1⊥...⊥ H_t.

[ Remark: e is odd since dim(V)=e· dim(W)=2t· dim(W)]

We proceed by induction on the composition length of V. 

If V has length 2 then it is trivial.

Suppose that the composition length is greater than 2. Let W be an irreducible submodule of V, and N a V-complement of W^⊥. By <cit.>, N is isomorphic to the dual of W, so it is isomorphic to W and totally isotropic . The sum W+N is direct, and the restriction of the form is non-singular.

In fact if w∈ W, n∈ N and ⟨ w+n,u⟩ =0 for all u∈ W+ N, in particular it holds ⟨ n,u⟩ =0 for all u∈ W, i.e n∈ W^⊥⋂ N=0.

So (W+ N)⋂ (W+ N)^⊥≤ W. If this intersection is W then in particular W≤ N^⊥, i.e N≤ W^⊥ that is a contradiction.

So H_1=W⊕ N is an hyperbolic symplectic module. Now

    H_1⋂ H_1^⊥   =(W⊕ N)⋂ (W⊕ N)^⊥=(W⊕ N)⋂ W^⊥⋂ N^⊥
       =((W⊕ N)⋂ W^⊥)⋂ N^⊥=W⋂ N^⊥=0.

Thus V=H_1⊥ H_1^⊥. Now H_1^⊥ is symplectic, since (H_1^⊥)^⊥=H_1. In particular it is non irreducible and so it has composition length greater than 1 and we can use the inductive hypothesis.


Let V be a symplectic, faithful 𝔽_3G-module that is a direct sum of irreducible anisotropic modules and dim(V)=2n. If | G| is odd, O_3(G)=1 and P∈ Syl_3(G) we have j=log_3(| P|)≤β_3(2/3n).


We proceed by induction on n, let V be a completely reducible, faithful, symplectic, 𝔽_3[G] module, with dimension 2n. 


Step 1:We can take V irreducible


Let V=W_1⊕ ...⊕ W_d with W_i irreducible anisotropic modules, with 2m_i=dim(W_i). So if C_i=C_G(W_i), the W_is are irreducible, faithful, symplectic G/C_i modules, so by inductive hypothesis we have 

    log_3(| G/C_i|_3)≤β_3(2/3m_i).

Now ⋂ C_i=1, so G is isomorphic to a subgroup of G/C_1× ...× G/C_d and thus

    log_3(| G|_3)≤∑β_3(2/3m_i)≤β_3(2/3∑ m_i)≤β_3(2/3n)



Step 2: Suppose that V is imprimitive


Let N G maximal such that V_N is not homogeneous, 

    V_N=e(V_1⊕...⊕ V_t)=W_1⊕...⊕ W_d,

where the W_is are the homogeneous components. Then S=G/N permutes faithfully and primitively the homogeneous components of V_N. Now the homogeneous components are symplectic with respect to the induced form, and are all orthogonal to each other, by <cit.> and <cit.>.

Now suppose that an homogeneous component contains an irreducible anisotropic submodule. Then since the homogeneous components are conjugated each of them contains an irreducible anisotropic submodule. Thus by using <cit.> on each homogeneous component we have that V is direct sum of simple non singular submodules. We can then apply the inductive hypothesis on each W_i, and if N_i=N/C_N(W_i) we have

    log_3(| N_i|_3)≤β_3(2/3m).

Now doing as in step 1 we get

    log_3(| N|_3)≤ dβ_3(2/3 m).

Since S is isomorphic to a subgroup of S_d, 

    log_3(| G|_3)≤ dβ_3(2/3 m)+λ_3(d)≤β_3(2/3 md)≤β_3(2/3 n).

If no homogeneous component contains an irreducible anisotropic submodule then by <ref> we have that e is even but by <ref> e must be odd, so we have a contradiction.


Step 3: Suppose finally that V is primitive


If V is primitive then every normal abelian subgroup of G is cyclic, F(G)=EU with U=Z(F(G)) cyclic, all the Sylow subgroups of E are extraspecial or have prime order and E⋂ U=Z=Soc(U).


If E=Z, then the Fitting subgroup of G is cyclic and since G is solvable we know that C_G(F(G))≤ F(G), i.e C_G(F(G))=F(G). Now V_F(G) is homogeneous, in particular simple by <cit.>.

So by <cit.> V can be identified with 𝔽_3^2n^+ and G with a subgroup of the semidirect product between 𝔽_3^2n^× and Gal(𝔽_3^2n|𝔽_3), with G⋂𝔽_3^2n^×=F(G). But then | G/F(G)| is cyclic and its order divides 2n. However | G| is odd, so | G/F(G)|≤ n=dim(V)/2, thus

    log_3(| G/F(G)|_3)≤λ_3(n)=β_3(2/3n).



We can then suppose E> Z. Then E/Z=Q_1/Z× ...× Q_k/Z, and e^2=| E:Z|=q_1^2m_1...q_k^2m_k. Now if A=C_G(Z) then A/C_A(Q_i/Z)≲ Sp(2m_i,q) and if we consider C_i=C_A(Q_i/Z), then by <cit.> we have

    log_3(| A:C_i|_3)≤β_3(q_i^m_i).

Now we have that C=⋂ C_i=F(G).

In fact E≤ C and U≤ C imply F(G)≤ C. U is the centralizer of E in G, thus C/U acts faithfully on E, centralizing both E/Z and Z. But E/Z is abelian, so | C/U| divides | E:Z| by <cit.>, thus C/U=E/Z, i.e. C=F(G).

Now by <cit.>

    log_3(| A:F(G)|_3)≤∑log_3(| A:C_i|_3)≤β_3(∑ q_i^m_i).

On the other hand | G:A|≤(| Z|-1)/2, since G/A embeds in Aut(Z). Let then s be the maximum integer such that 3^s≤ (| Z|-1)/2, so we can write:

    log_3(| G:F(G)|_3)≤β_3(∑ q_i^m_i)+s.

Now from <cit.> we have that | V| = | W|^eb, where W is an irreducible U-module such that the action of U is fixed point free. Obviously | W|≥| U|-1. 

If b> 1 then | V|≥ (| U|-1)^2e. Otherwise V_U is sum of e irreducible submodules all isomorphic to W.

If there is an irreducible non singular submodule M≤ V_U then M≃ W and thus U embeds in Sp(2m,3), with 2m=dim(W). So by <cit.>

    | V|≥| W|≥( | U|-1)^2e.

If every irreducible submodule of V_U=eW is totally isotropic then we have a contradiction. In fact by <ref> e is even, but e|| G| so it would be odd.


We can then say that | V|≥ (| U|-1)^2e, and let t be the minimum integer such that 3^t≥ (| U|-1)^2, so 2n≥ te. We remark that | U|≥ 5, so t≥ 3.

If e is not a prime power then since a+b+c≤ abc/6 for a≥ 5,b≥ 7,c≥ 3, we have 

This is true in fact if

    a=3+t,  t≥ 2
    
    b=3+s,  s≥ 4
    
    c=3+k,  k≥ 0
 then 

    a+b+c≤ abc/6    sse  54+6t+6s+6k≤ (3+t)(3+s)(3+k)
        sse  54+6t+6s+6k≤ 27+ 9(t+s+k)+3(ts+sk+tk).

But this holds since 27+3ts+3t+3s≥ 24+12+6+27≥ 54.


    log_3(| G:F(G)|_3)   ≤β_3(∑ q_i^m_i)+s≤β_3(∑ q_i^m_i)+t/2≤β_3(∑ q_i^m_i)+β_3( t)
       ≤β_3(∑ q_i^m_i+t)≤β_3(et/6)≤β_3( 2/3n).

Otherwise if e=q^m is a prime power we observe that if q^m+t≥ 13 and t,q^m≥ 5 it holds q^m+t≤ q^mt/3=et/3, thus the theorem is proved in this case.

Now we only need to check a few cases, i.e. the pairs (t,q^m) that satisfy t+q^m< 13. In particular since e∈{ 5,7,11}, t cannot be greater than 8. Thus the cases we need to check are (t≤ 7,5),(t≤ 5,7), (1,11). Moreover since 3^t≥ (| U|-1)^2 we can exclude (1,11).


Case e=7, t≤ 5. Since 7| e we have that 7 is a prime divisor of | U|, so we can exclude the cases t≤ 3. If t=5, then 3^5≥ (| U|-1)^2≥ 3^4, i.e (| U|-1)/2≤ 8, so s=1. Since 2n≥ et≥ 28, then β_3(2/3n)≥β_3(28/3)=4+1 and keeping in mind β_3(7)=4 it follows 

    β_3(2/3n)≥ 5≥ s+β_3(7)≥ log_3(| G:F(G)|_3).



Case e=5, t≤ 7. Now 3^7≥ (| U|-1)^2 thus | U| can be equal to 5,25,35. Z is the socle of U so | Z|=35 or | Z|=5. In particular s=0 or s=2, then s≤ 2. If 2n≥ 35, i.e t=7, the following holds 

    β_3(2/3n)≥ 6≥ 4≥ 2+β_3(5)≥ s+β_3(e)≥ log_3(| G:F(G)|_3).

If t≤ 6 then (35-1)^2> 3^6, so | U| =5 or | U|=25. In particular s=0, then

    β_3(2n/3)≥ 2=β_3(e)+0≥ log_3(| G:F(G)|_3).





§ MAIN RESULTS

In this section we finally deal with the degrees of primitive characters. We remind the definition of a primitive character.

We say that a character χ is primitive if it cannot be induced from any proper subgroup of G.

For the benefit of the reader we state a couple of results that will be used.

Let V be a faithful 𝔽_p[G]-module. with G p-solvable. Assume further that O_p(G)=1 and let Q be a Sylow p-subgroup of G. Then | Q|≤| V|^α/√(p), where 

    α=
    
    p/(p-1)^2     if p is a Fermat prime 
    
    1/(p-1)     otherwise
 



Let V be a completely reducible 𝔽[G]-module, where 𝔽 is any field. Suppose that C_V(G)=0 and let p be the smallest prime divisor of | G|. Then there exists g∈ G such that 

    dimC_V(g)≤1/p dimV.


We also state the definition of p-special characters. A comprehensive discussion of this subject can be found in <cit.> or in the more recent <cit.>.

Let π be a set of primes, G a π-separable group. We say that a character χ∈ Irr(G) is π-special if χ(1) is a π-number and o(θ) is a π-number for every irreducible constituent θ of the restriction χ_S for every subnormal subgroup S G. If π={ p} we say that χ is a p-special character.


Let χ be an irreducible primitive character of an odd order group G, p a prime dividing | G|. Then there is a g∈ G such that (χ(1))_p divides | cl_G(g)|_p.

 Since, by <cit.>, primitive characters are factorizable as a product of a (primitive)p-special character and a (primitive)p'-special character we can assume that χ is a p-special character. We can also assume that χ is faithful and so O_p'(G)=1 and F(G)=O_p(G).

Now since G has a primitive faithful character, every normal abelian subgroup of G is cyclic and central, thus there is a characterization of the Fitting subgroup: F(G)=EZ(G) is the central product of E, an extraspecial p-group and the centre of the group. Moreover V=E/Z(E) is a completely reducible, faithful, symplectic 𝔽_pG/F(G) module.

Now if P is a Sylow p-subgroup of G, the restriction map χ↦χ_P maps injectively the p-special characters of G to the irreducible characters of P. Thus

    χ(1)|| P:Z(P)|^1/2.

Moreover Z(P)=Z(F(G))=Z(G), in fact since G is solvable Z(P)⊆ C_G(F(G))⊆ Z(G)⊆ O_p⊆ P.


Now by (<ref>), χ(1) divides 

    | P:F(G)|^1/2| F(G):Z(G)|^1/2=| P:F(G)|^1/2| E:Z|^1/2.

We observe that G acts fixed point freely on V.

In fact if xZ∈ C_V(G), x∈ E, then x^gZ=xZ for all g∈ G. Thus ⟨ x⟩ Z is a normal subgroup of G. In particular ⟨ x⟩ Z is also abelian, thus cyclic and central, that implies x∈ Z and xZ=Z. 

Now if we use Theorem <ref> with G:=G/F(G), V=E/Z(E) and Q a p-Sylow subgroup of G/F(G), then | Q|=| P:F(G)|, and for p≥ 5 | P:F(G)|≤| V|^1/3. Thus χ(1)≤| E:Z|^1/6+1/2=| E:Z|^2/3. Since the minimum prime that divides | G| is at least 3, if g whose existence is guaranteed by Theorem <ref>, we have | cl_G(g)|_p≥ V^2/3≥χ(1).


We only need to prove the Theorem in the case p=3. Now the module V is the direct sum of irreducible anisotropic modules, that inherit the symplectic form induced by the commutator.

In fact if an irreducible module N/Z were totally isotropic then for all a,e∈ N we would have ⟨ aZ,eZ⟩ =0, i.e. [a,e]=1. Thus N would be abelian, but then it would be central and thus N≤ E⋂ Z(G)=Z.

So we may apply Theorem <ref> to G/F(G) and V, thus we have 

    | G:F(G)|_3≤ 3^β_3( 2/3n)=3^β_3(dim(V)/3)≤ 3^dim(V)/4=| V|^1/4

and then 

    χ(1)≤| V|^7/12≤| V|^2/3≤| cl_G(g)|_3



If | G|=p^aq^b is odd, and G has a normal or abelian Sylow subgroup, then for all primitive characters χ∈ Irr(G) there is g∈ G such that χ(1) divides | g^G|.


Let χ=αβ be a factorization of χ as a product of respectively a p-special character and a q-special. If G has a normal Sylow p-subgroup P, then | G:P|=q^b and β_P is irreducible, hence is linear and χ(1) is a prime power.

If G has an abelian Sylow p-subgroup then p∤χ(1), and then χ(1) is again a prime power.

We can now apply Theorem <ref>.

The following is an easy corollary of Theorem <ref> in the case V is a module of mixed characteristic.

Let V be  a finite G-module of mixed characteristic, G odd order group, V=P̂_1⊕P̂_2, with P̂_i non trivial 𝔽_p_iG-modules and π(| G|)={ p_1,p_2 }. If K G, C_V(K)=0 and V is K-completely reducible, then every coset of K contains a t such that dim(C_P̂_i(t))≤1/p_idim(P̂_i).


We use induction on dim(V). We can suppose that G=K⟨ t⟩ and that K acts faithfully. In fact if K does not act faithfully, then we may consider K/C_K(V) G/C_K(V).

If G/C_K(V) is a p_i-group, then K/C_K(V) is nilpotent and thus acts trivially on P̂_i, since P̂_i is K-completely reducible.

In fact consider P̂_i as an abelian group and let H=P̂_i⋊ P, where P is the Sylow p_i-subgroup of K/C_K(V). W being irreducible means that it is a minimal normal subgroup of H, but H is a p_i-group, so W is central and has order p_i, i.e. W has dimension 1 and the action of P_i is trivial.

But C_V(K)=0 implies that P̂_i is trivial, and this is a contradiction. So it must be that π(G/C_K(V))=π(G) and we are done by induction.

Suppose then that K acts faithfully and consider N≤ K minimal normal subgroup of G. If U=C_V(N)>0, then U< V and U admits the action of G, in fact if v∈ U and x∈ N, then

    v^gx=v^x^-1gx=v^[x,g^-1]g=v^x'g=v^g.

So V=U⊕ W with [V,N]=W is a sum of G-modules. In fact if W is the N-complement of U, [V,N]⊆ W. But by the N-complete reducibility [V,N] has a complement, that must be N-invariant and thus W=[V,N].


We remark that C_U(K)=0. Now if, say, P̂_1⋂ U=0 then by Theorem <ref> we have that ∃ x∈ Kt such that

    dim(C_P̂_i⋂ U(x))≤1/pdim(P̂_i⋂ U).
 
Suppose then now that P̂_i∩ U≠ 0 for all the is. If K is a p_j-group, then K is nilpotent and so by comprimality as before we would have P̂_j⋂ U=0, but this is a contradiction. 

Then we must have that π(K)={ p_1,p_2}. So by induction ∃ x∈ Kt such that 

    dim(C_P̂_i⋂ U(x))≤1/pdim(P̂_i⋂ U).
 
Now C_W(N)=0 and we can suppose that N is an elementary abelian p_1-group, but W⋂P̂_1 is a faithful completely reducible N/C_N(W⋂P̂_1)-module, and thus C_N(W⋂P̂_1)=N, i.e. W⋂P̂_1=0. Then by Theorem <ref> there is y∈ Nx⊆ Kt such that 

    dim(C_P̂_i⋂ W(y))≤1/pdim(P̂_i⋂ W).
 
But y=nx, so C_U(y)=C_U(x) and then we are done.


If C_V(N)=0 we can replace K with N, and suppose that it is an elementary abelian p_1-group. But then P̂_1 is a faithful completely reducible K/C_K(P̂_1)-module, so C_K(P̂_1)=K, i.e. K acts trivially on P̂_1 and thus P̂_1=0 but this is a contradiction. 

We want to remark that the Proposition <ref>, without any restrictions on the prime factors of the order of G, is false, as we can see from the following example. 

Take G=Kt=⟨ k⟩⟨ t⟩ isomorphic to the Klein group. Take then p_1,p_2 odd prime numbers and S_1,T_1 the linear 𝔽_p_1G, 𝔽_p_2G-modules, such that if R_S is the representation associated to S_1, R_1:G→𝔽_p_1^*, then Ker(R_S)=⟨ kt⟩ and as well Ker(R_T)=⟨ kt⟩. Now take S_2,T_2 linear 𝔽_p_1G,𝔽_p_2G-modules such that the kernel of the associated representation is ⟨ t⟩. 

Consider then:

    V_1=2S_1⊕ S_2
    
    V_2=T_1⊕ 2S_2

These two modules have dimension 3 obviously. If we look at the fixed points of the coset Kt then

    dimC_V_1(kt)=2> 3/2
    
    dimC_V_2(t)=2> 3/2,
 
so there is no element g in the coset Kt such that dimC_V_i(g)≤1/2dim(V_i).

The following is the only case where we were able to prove that the degree of a primitive character divides the size of a conjugacy class, assuming that the degree is not necessarily a prime power.


Let G be a meta-nilpotent group, | G| odd, π(G)={ p_1,p_2}, χ∈Irr(G) primitive character. Then there is g∈ G such that χ(1) divides | cl_G(g)|.
 

We can take χ faithful, and so we have again the following structure of the Fitting subgroup, F(G)=EZ(G). We then assume that χ(1)=p_1^ap_2^b, with a,b≥ 1, otherwise the thesis immediately follows from Proposition <ref>. 

Let V=E/Z, V=P̂_1⊕P̂_2, with P̂_i 𝔽_p_i-modules, even trivial. Let χ=α_1α_2 be a factorization of χ as a product of p_i-specials characters. Notice that if G_i/F(G) are the Sylow p_i-subgroups of G/F(G) then by <cit.> we have that (α_i)_G_i+1 is primitive and irreducible, where the indices are modulo 2.

But then if O_p_i:=O_p_i(G) we have that (α_i)_O_p_i is irreducible, and then α_i(1) divides | O_p_i:Z(O_p_i)|^1/2. That is χ(1) divides | V|.
 Now if, say, P̂_1 is trivial then O_p_1 is central and thus α_1 is linear, but we have already excluded this occurrence. So we can take P̂_i non trivial modules, and then by Proposition (<ref>) there is an element g∈ G such that 

    | g^G|_p_i≥| E:Z|_p_i^2/3≥| O_p_i:Z(O_p_i)|^2/3≥α_i(1),

thus | g^G| is divided by χ(1).


I would like to thank Silvio Dolfi, for many useful discussions and comments, and Charles Eaton, for a careful reading of the manuscript.

siam

