4 amended text

A good potential supply of this sort of character is provided by Frobenius groups. These are finite groups $G$ of the form $G = KH$, where $K \cap H = 1$ and $K \lhd G$, and furthermore $C_{G}(x) \leq K$ for all non-identity elements $x \in K$. By a Theorem of Thompson the group $K$ is necessarily nilpotent.

One Frobenius group not mentioned in your examples is given by $H \cong {\rm SL}(2,5)$ and $K$ elementary Abelian of order $121$, which admits a regular action (on non-identity elements) by $H$. The semidirect product gives an example of a finite group $G$ which is not solvable, but has an irreducible character of quite large degree. Frobenius complements which are not solvable are very rare though.

In general, the irreducible characters $\chi$ of a (general) Frobenius group $G$ which do not contain $K$ in their kernels have degrees of the form $|H|\mu(1)$, where $\mu$ is an irreducible character of $K$. For such a $\chi$ to have quite large degree, we clearly need $|H|\mu(1)^2 \geq 0.64 |K|$. Notice also that $|K| \equiv 1$ (mod $|H|$). If $|H| < |K|-1$, then we obtain the contradiction $\mu(1)^2 > 1.28 |K|$. Thus the only Frobenius groups with irreducible characters of quite large degree are those in which the complement $H$ acts transitively on non-identity elements of the kernel $K$. This forces the kernel $K$ to be elementary Abelian. Finite vector spaces which admit a group of automorphisms transitive on non-zero elements are probably classified somewhere (C. Hering?). Anyway, it looks as though you won't get many new examples from Frobenius groups (later edit-as is made explicit by Noah Snyder's comment below. Previous version did not say what I meant in any case).

3 typo

A good potential supply of this sort of character is provided by Frobenius groups. These are finite groups $G$ of the form $G = KH$, where $K \cap H = 1$ and $K \lhd G$, and furthermore $C_{G}(x) \leq K$ for all non-identity elements $x \in K$. By a Theorem of Thompson the group $K$ is necessarily nilpotent.

One Frobenius group not mentioned in your examples is given by $H \cong {\rm SL}(2,5)$ and $K$ elementary Abelian of order $121$, which admits a regular action (on non-identity elements) by $H$. The semidirect product gives an example of a finite group $G$ which is not solvable, but has an irreducible character of quite large degree. Frobenius complements which are not solvable are very rare though.

In general, the irreducible characters $\chi$ of a (general) Frobenius group $G$ which do not contain $K$ in their kernels have degress degrees of the form $|H|\mu(1)$, where $\mu$ is an irreducible character of $K$. For such a $\chi$ to have quite large degree, we clearly need $|H|\mu(1)^2 \geq 0.64 |K|$. Notice also that $|K| \equiv 1$ (mod $|H|$). If $|H| < |K|-1$, then we obtain the contradiction $\mu(1)^2 > 1.28 |K|$. Thus the only Frobenius groups with irreducible characters of quite large degree are those in which the complement $H$ acts transitively on non-identity elements of the kernel $K$. This forces the kernel $K$ to be elementary Abelian. Finite vector spaces which admit a group of automorphisms transitive on non-zero elements are probably classified somewhere (C. Hering?). Anyway, it looks as though you won't get many new examples from Frobenius groups.

2 minor amendment of text

A good potential supply of this sort of character is provided by Frobenius groups. These are finite groups $G$ of the form $G = KH$, where $K \cap H = 1$ and $K \lhd G$, and furthermore $C_{G}(x) \leq K$ for all non-identity elements $x \in K$. By a Theorem of Thompson the group $K$ is necessarily nilpotent.

One Frobenius group not mentioned in your examples is given by $H \cong {\rm SL}(2,5)$ and $K$ elementary Abelian of order $121$, which admits a regular action (on non-identity elements) by $H$. The semidirect product gives an example of a finite group $G$ which is not solvable, but has an irreducible character of quite large degree. Frobenius complements which are not solvable are very rare though.

In general, the irreducible characters $\chi$ of a (general) Frobenius group $G$ which do not contain $K$ in their kernels have degress of the form $|H|\mu(1)$, where $\mu$ is an irreducible character of $K$. For such a $\chi$ to have quite large degree, we clearly need $|H|\mu(1)^2 \geq 0.64 |K|$. Notice also that $|K| \equiv 1$ (mod $|H|$). If $|H| < |K|-1$, then we obtain the contradiction $\mu(1)^2 > 1.28 |K|$. Thus the only Frobenius groups with irreducible characters of quite large degree are those in which the complement $H$ acts transitively on non-identity elements of the kernel $K$. This forces the kernel $K$ to be elementary Abelian. Finite vector spaces which admit a group of automorphisms transitive on non-zero elements are probably classified somewhere (C. Hering?). Anyway, it looks as though you won't get many new examples from Frobenius groups.

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