Groups with an irreducible character of large degree are solvable

TLDR: Theorem 4.5 as mentioned in this paper states that a finite group G admits an irreducible complex character of large degree if t(1)2 = [G:A ]2 =[G:Z], that is whenever the two upper bounds noted above are achieved simultaneously.

Abstract: The degree of an irreducible complex character afforded by a finite group is bounded above by the index of an abelian normal subgroup and by the square root of the index of the center. Whenever a finite group affords an irreducible character whose degree achieves these two upper bounds the group must be solvable. Let G be a finite group with an irreducible (complex) character ?. If Z is the center of G it is easy to prove that ~(1)2_ [G:Z] and if A is an abelian normal subgroup of G it is easy to show that t(1) ? [G:A A (see pp. 364-365 of [1]). Say the group G admits an irreducible character of large degree if t(1)2 = [G:A ]2 = [G:Z], that is whenever the two bounds noted above are achieved simultaneously. Such groups arise in the theory of projective representations and the Galois theory in general rings [2]. The purpose of this note is to give proof of the result stated in the title, thus verifying a special case of a conjecture in [2]. Througlhout all unexplained terminology is as in [1]. THEOREM. Let G be a group with center Z and abe/ian normal subgroup A so that there is an irreducible complex character v on G with t(1) 2 = [G: A ]j2= [G: Z]. Then G is solvable. PROOF. A theorem of P. Hall (Theorem 4.5, p. 233 of [31]) asserts that a group is solvable if and only if every p-sylow subgroup has a complement. This theorem will be applied to G/A to give the result. Since the degree of any irreducible character is bounded by the index of an abelian subgroup, A is a maximal abelian normal subgroup of G, so ZCA. If 7r, II, and P are sylow p-subgroups of Z, A, and G respectively then 7rCIICP. M\oreover wr is contained in the center of P, II is an abelian normal subgroup of P, and II is a normal subgroup of G. Arguing as in [2 ] we show p e = mX where X is irreducible on P and X(1) = [P:ll]. By Schur's lemma we have tJ z = r(1)q for some linear character $ of Z. Then (D, f G)(P Z, k) = (1) so by counting degrees D(1)v=OG. Let R be the subgroup of G generated by Z and P, and let X be an irreducible character of R contained in q1?. By Schur's Received by the editors December 29, 1969. AMS Subject Classifications. Primary 2080; Secondary 2027, 1670.