On the minimal norm of a non-regular generalized character of an arbitrary finite group Geoffrey R. Robinson Department of Mathematics University of Aberdeen Aberdeen AB24 3UE Scotland United Kingdom e-mail:g.r.robinson@abdn.ac.uk

INTRODUCTION AND NOTATION: As usual, for G a finite group, we denote the set of non-identity elements of G by G^{#}, the number of conjugacy classes of G by k(G) and the set of complex irreducible characters of G by Irr(G). When H is a non-trivial subgroup of G, we let k_{+}(G, H) denote the number of irreducible characters of G which do not vanish identically on H^{#} (and we remark that k_{+}(G, H) = k(G) when |H| ≥ \sqrt{|G|}).

For θ a generalized character of a finite group G, we set

σ(G^{#}, θ) = ∑_{g∈G^{#}} |θ(g)|^{2}.

Notice that

σ(G^{#}, θ) = |G|⟨θ, θ⟩ − θ(1)^{2},

where ⟨, ⟩ is the usual inner product on the space of complex-valued class functions of G. We denote the regular character of G by ρ. We let cf(G) denote the space of complex-valued class functions of G, and for α, β ∈ cf(G), we set

(α, β) = |G|⟨α, β⟩ − α(1)β(1).

Then (,) induces an inner product on the quotient space cf(G)/ℂρ. We label the irreducible characters of G as μ_{1}(= 1), μ_{2}, …, μ_{k} where μ_{1}(1) ≤ … ≤ μ_{k}(1).

In the earlier papers [1],[2] and [3], we considered the minimum possible value of σ(G^{#}, θ) for θ a generalized character (other than a multiple of the regular character) of a non-trivial finite group G. We denote this minimum possible value by m(G) (which is clearly a positive integer). Note that with the above notation, this is the minimum non-zero value of (θ, θ) for θ ∈ ℤIrr(G)\ℤρ.

We proved in [1] that m(G) ≥ k(G) − 1 for every non-trivial finite group G. In the papers [2],[3] we proved that m(G) = |G| − μ_{k}(1)^{2} whenever G is a non-trivial nilpotent group. In particular, in that case,

m(G) ≥ |G| − [G : Z(G)]

whenever G since μ_{k}(1)^{2} divides [G : Z(G)].

We asked in [3] whether it was the case that for every non-trivial finite group G we have

m(G) ≥ \frac{|G|}{μ_{k}(1)} − 1

(which would yield m(G) > \sqrt{|G|} − 1) . We answer that question in the affirmative here. Notice that if G is a Frobenius group with Abelian kernel K and complement H, then m(G) = |K| − 1. In that case, the maximal degree of an irreducible character of G is |H|, and we do have

m(G) = \frac{|G|}{μ_{k}(1)} − 1.

For if θ is a generalized character of G which does not vanish identically on H^{#}, we see that σ(G^{#}, θ) ≥ |K|σ(H^{#}, Res_{H}^{G}(θ)) ≥ |K|, as there are |K| conjugates of H which intersect trivially with each other. On the other hand, if the generalized character θ does not vanish identically on K^{#}, then

σ(G^{#}, θ) ≥ m(K) = |K| − 1,

as K is Abelian. Finally, let θ be the sum of the distinct irreducible characters of G which do not contain K in their kernels. By the orthogonality relations, we see that θ(x) = −1 for all x ∈ K^{#}, while also θ vanishes identically outside K, so that σ(G^{#}, θ) = |K| − 1, as claimed.

In view of these examples, our main theorem may be regarded as optimal in this generality. We prove:

THEOREM 1 : Let G be a non-trivial finite group. Then

m(G) ≥ \frac{|G|}{μ_{k}(1)} − 1.

Furthermore, equality is attained if and only if G is either Abelian, or else is a Frobenius group with an Abelian Frobenius kernel.

As noted above, Theorem 1 has an interpretation in terms of the minimal non-zero value of a certain semi-definite integral quadratic form.

COROLLARY 2: Let G be a non-trivial finite group and let a_{1}, a_{2}, …, a_{k} be any k integers such that for some i ≠ j, we have a_{i}μ_{j}(1) ≠ a_{j}μ_{i}(1). Then

∑_{i<j}(a_{i}μ_{j}(1) − a_{j}μ_{i}(1))^{2} ≥ \frac{|G|}{μ_{k}(1)} − 1.

COROLLARY 3: Let H be a non-trivial subgroup of the finite group G and let μ be a complex irreducible character of maximal degree of H. Then we have

k_{+}(G, H) ≤ \frac{μ(1)|H| − μ(1)}{|H| − μ(1)}max_{h∈H^{#}}|C_{G}(h)|.

PROOF OF THEOREM 1

Suppose that m(G) < \frac{|G|}{μ_{k}(1)}. To prove the Theorem, by the cases discussed above, we may suppose that G is non-Abelian. Let us choose a generalized character θ of G such that m(G) = σ(G^{#}, θ). We will prove that θ(x) ∈ {0, −1} for all x ∈ G^{#}, and that we may suppose that θ is a multiplicity free irreducible character of G whose irreducible constituents all have degree μ_{k}(1), and that if there are r such constituents, then |G| = μ_{k}(1) + rμ_{k}(1)^{2}.

If θ has the form tρ + δμ_{i} for some integer t, some sign δ and some i, then we may replace θ by μ_{i}. Then we have

σ(G^{#}, θ) = |G| − μ_{i}(1)^{2},

so we may suppose that i = k, as μ_{i} must also have maximal degree by the choice of θ. We then have

σ(G^{#}, θ) = |G| − μ_{k}(1)^{2} ≥ \frac{|G|}{μ_{k}(1)} − 1.

Furthermore, equality forces either μ_{k}(1) = 1 (in which case G is Abelian, contrary to hypothesis), or else |G| = μ_{k}(1)^{2} +μ_{k}(1). In the latter case, we have ∑_{i=1}^{k−1} μ_{i}(1)^{2} = μ_{k}(1). Now ⟨θ, μ_{1}⟩ = 0, so that ∑_{g∈G^{#}} μ_{k}(g) = −μ_{k}(1). Under the current assumptions, we also have

σ(G^{#}, θ) = μ_{k}(1).

Now μ_{k} is the unique irreducible character of G of its degree, so it is rationalvalued. Hence we must have μ_{k}(g) = −1 whenever μ_{k}(G) ≠ 0. In order to establish the initial claim, then, we may now suppose that θ does not have the form tρ + δμ_{i} for some integer t, some sign δ and some i.

By the arguments of [2] and [3], we may suppose that θ is a rational-valued character of G with 0 < θ(1) < \frac{|G|}{2} ( in which irreducible characters of the same degree occur with equal multiplicity), and that ⟨θ, μ_{i}⟩ is the closest integer to \frac{θ(1)μ(1)}{|G|} for each irreducible character μ_{i} of G. For the convenience of the reader, we review the necessary results using the notation currently adopted. After subtracting integer multiples of ρ from θ and replacing θ by −θ if necessary, we may suppose that 0 < θ(1) ≤ \frac{|G|}{2}.

For (by the choice of θ) we have

(θ − δμ_{i}, θ − δμ_{i}) ≥ (θ, θ)

for each i and any choice of sign δ. This leads easily to

|(θ, μ_{i})| ≤ \frac{|G| − μ_{i}(1)^{2}}{2}

and then

|⟨θ, μ_{i}⟩ − \frac{θ(1)μ_{i}(1)}{|G|}| ≤ \frac{1}{2} (1 − \frac{μ_{i}(1)^{2}}{|G|} ) .

Hence ⟨θ, μ_{i}⟩ is the (unique under current assumptions) closest integer to \frac{θ(1)μ_{i}(1)}{|G|},

and only depends on μ_{i}(1). Notice that we can’t now have θ(1) = \frac{|G|}{2}, for if we did, we would have the contradiction

|⟨θ, μ_{1}⟩ − \frac{1}{2}| < \frac{1}{2}.

Thus we have 0 < θ(1) < \frac{|G|}{2}. Also, θ is a rational (integer) valued character, since it contains algebraically conjugate irreducible characters with the same non-negative multiplicity.

Notice that the trivial character does not now occur as an irreducible constituent of θ. Hence we have ∑_{g∈G^{#}} θ(g) = −θ(1), so that

∑_{g∈G^{#}} |θ(g)| ≥ θ(1)

and certainly

σ(G^{#}, θ) ≥ θ(1)

( as θ is integer-valued).

Now let us write

θ = ∑_{i=1}^{k} a_{i}μ_{i},

and set

ϵ_{i} = a_{i} − \frac{θ(1)μ_{i}(1)}{|G|}.

Then we have

∑_{i=1}^{k} ϵ_{i}μ_{i}(1) = 0

and

σ(G^{#}, θ) = |G| (∑_{i=1}^{k} |ϵ_{i}|^{2})

( for notice that θ agrees with ∑_{i=1}^{k} ϵ_{i}μ_{i} on G^{#}). We have noted already that |ϵ_{i}| < \frac{1}{2} for each i.

Suppose that a_{i} ≥ 2 for some i. Then we have

\frac{θ(1)μ_{i}(1)}{|G|} > \frac{3}{2},

so that

σ(G^{#}, θ) ≥ θ(1) > \frac{G}{μ_{i}(1)} ≥ \frac{|G|}{μ_{k}(1)},

contrary to hypothesis. Hence we have a_{i} ∈ {0, 1} for each i.

Now suppose that ϵ_{i} < 0 for i = 1, …, k − r, and that ϵ_{i} > 0 for i = k − r + 1, …, k. Then

θ = ∑_{i=1}^{r} μ_{k−r+i}.

Since we have dealt with the case that θ is irreducible, we may, and do, suppose that r > 1.

By assumption, we have

\frac{|G|}{μ_{k}(1)} − 1 ≥ m(G) = σ(G^{#}, θ) ≥ θ(1),

while we clearly also have

θ(1) ≥ \frac{|G| − ∑_{i=1}^{k−r} μ_{i}(1)^{2}}{μ_{k}(1)} ,

so that

∑_{i=1}^{k−r} μ_{i}(1)^{2} ≥ μ_{k}(1)

( and hence

|G| ≥ μ_{k}(1) + ∑_{i=k−r+1}^{k} μ_{i}(1)^{2}).

Now we have

σ(G^{#}, θ) = ∑_{i<j}(a_{i}μ_{j}(1) − a_{j}μ_{i}(1))^{2} ≥ r (∑_{i=1}^{k−r} μ_{i}(1)^{2}) ,

using the fact that a_{j} = 1 for j > k − r and a_{i} = 0 for i ≤ k − r.

Thus we have (by assumption)

r [|G| − ( ∑_{i=k−r+1}^{k} μ_{i}(1)^{2})] ≤ σ(G^{#}, θ) ≤ \frac{|G|}{μ_{k}(1)} − 1,

so that

|G| (1 − \frac{1}{rμ_{k}(1)}) < ( ∑_{i=k−r+1}^{k} μ_{i}(1)^{2}) .

Now for n ≥ 2, we have

\frac{∑_{i=k−r+1}^{k} μ_{i}(1)^{2}}{r^{n}μ_{k}(1)^{n}} ≤ \frac{1}{r^{n−1}μ_{k}(1)^{n−2}} .

Hence

|G| < ∑_{i=k−r+1}^{k} μ_{i}(1)^{2} + \frac{∑_{i=k−r+1}^{k} μ_{i}(1)^{2}}{rμ_{k}(1)} + \frac{1}{r} (1 − \frac{1}{rμ_{k}(1)})^{−1}

= ∑_{i=k−r+1}^{k} μ_{i}(1)^{2} + \frac{∑_{i=k−r+1}^{k} μ_{i}(1)^{2}}{rμ_{k}(1)} + \frac{1}{r − \frac{1}{μ_{k}(1)}} .

Now r > 1 and also μ_{k}(1) > 1 (as G is non-Abelian).

Thus

|G| < 1 + μ_{k}(1) + ∑_{i=k−r+1}^{k} μ_{i}(1)^{2},

so that we now have

|G| = μ_{k}(1) + ∑_{i=k−r+1}^{k} μ_{i}(1)^{2},

as we already have

|G| ≥ μ_{k}(1) + ∑_{i=k−r+1}^{k} μ_{i}(1)^{2}.

We also now have

\frac{|G|}{μ_{k}(1)} − 1 ≤ θ(1) ≤ σ(G^{#}, θ) ≤ \frac{|G|}{μ_{k}(1)} − 1,

so that

θ(1) = σ(G^{#}, θ) = \frac{|G|}{μ_{k}(1)} − 1.

Furthermore, we have μ_{i}(1) = μ_{k}(1) whenever i > k − r. Hence |G| = μ_{k}(1) + rμ_{k}(1)^{2}.

Furthermore, we also have |θ(x)|^{2} = 1 = −θ(x) for all x ∈ G^{#} such that θ(x) ≠ 0. Thus the claim of the first paragraph of the proof is complete, and in the discussion which follows, we allow the possibility that r = 1.

The orthogonality relations (and the fact that μ_{k}(1) = ∑_{i=1}^{k−r} μ_{i}(1)^{2}) now yield that x ∈ ⋂_{i=1}^{k−r} kerμ_{i} whenever θ(x) ≠ 0. Conversely, if θ(x) = 0, we have ∑_{i=1}^{k−r} μ_{i}(1)μ_{i}(x) = 0.

Set K = ⋂_{i=1}^{k−r} kerμ_{i}. Now

|G| = μ_{k}(1) + rμ_{k}(1)^{2}

and

|K| = 1 + θ(1) = 1 + rμ_{k}(1).

Thus K is a Hall subgroup of G, and there is a complement, H say, to K in G. Since μ_{k}(1) = |H|, we see that, for alli > k − r, we have μ_{i}(y) = 0 whenever gcd(|⟨y⟩|, |H|) > 1. In particular (using a well-known observation of Feit and Thompson),

|C_{G}(y)| = |C_{G/K}(yK)| = |C_{H}(y)|

for all y ∈ H^{#}. Thus C_{G}(y) ≤ H for all y ∈ H^{#} and G is a Frobenius group with kernel K, complement H. Furthermore, every irreducible character of G which does not contain K in its kernel has degree μ_{k}(1) = [G : K], so that every non-trivial irreducible character of K is linear. Hence K is Abelian.

PROOF OF COROLLARY 2: As noted in [1], if θ = ∑_{i=1}^{k} a_{i}μ_{i}, then (θ, θ) = ∑_{i<j}(a_{i}μ_{j}(1) − a_{j}μ_{i}(1))^{2}.

PROOF OF COROLLARY 3: By the orthogonality relations for G and Theorem 1 ( applied to H), and noting the presence of the trivial character, we have

∑_{h∈H^{#}} |C_{G}(h)| ≥ (|H| − 1) + (k_{+}(G, H) − 1) (\frac{|H|}{μ(1)} − 1) .

REFERENCES

[1] Robinson,G.R., A bound on norms of generalized characters with applications, Journal of Algebra, 212,(1999),660-668.

[2] Robinson,G.R., More bounds on norms of generalized characters with applications to p-local bounds and blocks, Bulletin of London Mathematical Society, 37,(4),(2005),555-565.

[3] Robinson,G.R., On generalized characters of nilpotent groups, Journal of Algebra,308,(20007), 822-827.