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Geoff Robinson
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As Will Sawin suggests, the answer to the last version of the question is negative. If we tak $G$ to be a non-Abelian finite simple group which is not a doubly transitive permutation group, and $\chi$ to be a non-trivial complex irreducible character of last degree o $G,$ then it is impossible to write $chi$$\chi$ + any multiple of tehthe trivial character as a sum of characters each induced from linear characters of (not necssarilynecessarily proper) subgroups. For if $\lambda$ is a non-trivial linear character of a proper subgroup $H$ of $G,$ then ${\rm Ind}_{H}^{G}(\lambda)$ does not contain the trivial character by Frobenius reciprocity. Neither can it be $\chi$, for otherwise ${\rm Ind}_{H}^{G}(1)$ would have a trivial constituent, and at least one non-trivial irreducible constituent $\mu$. But then $\mu(1) < \chi(1)$, contrary to the choice of $\chi.$ On the other hand,if $H$ is a proper subgroup of $G$, the permutation character ${\rm Ind}_{H}^{G}(1)$ only contains the trivial character once, but it can't be $1 + \chi,$ for otherwise $G$ would be a doubly transitive permutation character on the cosets of $H$ in $G,$ contrary to the choice of $G.$

As Will Sawin suggests, the answer to the last version of the question is negative. If we tak $G$ to be a non-Abelian finite simple group which is not a doubly transitive permutation group, and $\chi$ to be a non-trivial complex irreducible character of last degree o $G,$ then it is impossible to write $chi$ + any multiple of teh trivial character as a sum of characters each induced from linear characters of (not necssarily proper) subgroups.

As Will Sawin suggests, the answer to the last version of the question is negative. If we tak $G$ to be a non-Abelian finite simple group which is not a doubly transitive permutation group, and $\chi$ to be a non-trivial complex irreducible character of last degree o $G,$ then it is impossible to write $\chi$ + any multiple of the trivial character as a sum of characters each induced from linear characters of (not necessarily proper) subgroups. For if $\lambda$ is a non-trivial linear character of a proper subgroup $H$ of $G,$ then ${\rm Ind}_{H}^{G}(\lambda)$ does not contain the trivial character by Frobenius reciprocity. Neither can it be $\chi$, for otherwise ${\rm Ind}_{H}^{G}(1)$ would have a trivial constituent, and at least one non-trivial irreducible constituent $\mu$. But then $\mu(1) < \chi(1)$, contrary to the choice of $\chi.$ On the other hand,if $H$ is a proper subgroup of $G$, the permutation character ${\rm Ind}_{H}^{G}(1)$ only contains the trivial character once, but it can't be $1 + \chi,$ for otherwise $G$ would be a doubly transitive permutation character on the cosets of $H$ in $G,$ contrary to the choice of $G.$

Source Link
Geoff Robinson
  • 44.4k
  • 5
  • 123
  • 169

As Will Sawin suggests, the answer to the last version of the question is negative. If we tak $G$ to be a non-Abelian finite simple group which is not a doubly transitive permutation group, and $\chi$ to be a non-trivial complex irreducible character of last degree o $G,$ then it is impossible to write $chi$ + any multiple of teh trivial character as a sum of characters each induced from linear characters of (not necssarily proper) subgroups.