2013
DOI: 10.1016/j.jalgebra.2012.12.024
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Groups whose prime graphs have no triangles

Abstract: Let G be a finite group and let cd(G) be the set of all complex irreducible character degrees of G Let \rho(G) be the set of all primes which divide some character degree of G. The prime graph \Delta(G) attached to G is a graph whose vertex set is \rho(G) and there is an edge between two distinct primes u and v if and only if the product uv divides some character degree of G. In this paper, we show that if G is a finite group whose prime graph \Delta(G) has no triangles, then \Delta(G) has at most 5 vertices. … Show more

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Cited by 26 publications
(28 citation statements)
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“…By Lemma 3.2, r ∈ ρ(N ) and hence r | θ(1) for some θ ∈ Irr(N ). Since σ(M ) ≤ σ(G) ≤ 2 and M/N ∼ = S, by [T,Lemma 4.2] we deduce that θ extends to θ 0 ∈ Irr(M ). Now let λ ∈ Irr(M/N ) with 2 | λ(1).…”
Section: Proof Of Theorem Bmentioning
confidence: 79%
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“…By Lemma 3.2, r ∈ ρ(N ) and hence r | θ(1) for some θ ∈ Irr(N ). Since σ(M ) ≤ σ(G) ≤ 2 and M/N ∼ = S, by [T,Lemma 4.2] we deduce that θ extends to θ 0 ∈ Irr(M ). Now let λ ∈ Irr(M/N ) with 2 | λ(1).…”
Section: Proof Of Theorem Bmentioning
confidence: 79%
“…By Lemma 3.2, r | θ(1) for some θ ∈ Irr(N ). Since σ(M ) ≤ 2, by applying [T,Lemma 4.2], we deduce that θ extends to θ 0 ∈ Irr(M ). Using Gallagher's Theorem, we must have that σ(M/N ) = 1 and hence M/N ∼ = PSL 2 (4) or PSL 2 (8).…”
Section: Proof Of Theorem Bmentioning
confidence: 94%
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