11.8 (1990)

For a finite group $X$, let $\chi_1(X)$ denote the totality of the degrees of all irreducible complex characters of $X$ with allowance for their multiplicities. Suppose that $\chi_1(G) = \chi_1(H)$ for groups $G$ and $H$. Clearly, then $|G| = |H|$.
$\qquad$ a) Is it true that $H$ is simple if $G$ is simple?
$\qquad$ b) Is it true that $H$ is soluble if $G$ is soluble?
It is known that $H$ is a Frobenius group if $G$ is a Frobenius group.