Let \(G\) be a finite group and \(g\in G\). Suppose \(\chi(g)\in\RR\) for every complex irreducible character \(\chi\) of \(G\). What can be said about \(g\)?
Solution
One has \[ \chi(g^{-1})=\overline{\chi(g)} \] for every complex character \(\chi\). Indeed, a finite-group representation may be chosen unitary, so the eigenvalues of \(\rho(g)\) lie on the unit circle and the trace of \(\rho(g^{-1})\) is the complex conjugate of the trace of \(\rho(g)\).
By hypothesis \(\chi(g)\) is real for every irreducible character, hence \[ \chi(g^{-1})=\chi(g) \] for every irreducible \(\chi\).
Irreducible characters form a basis of the space of complex class functions on \(G\). Therefore two elements on which all irreducible characters take the same values must lie in the same conjugacy class. Hence \[ g^{-1}\sim g. \]
Conversely, if \(g\) is conjugate to \(g^{-1}\), then every character satisfies \[ \chi(g)=\chi(g^{-1})=\overline{\chi(g)}, \] so \(\chi(g)\) is real.
Thus \[ \boxed{\chi(g)\in\RR\text{ for every irreducible }\chi \iff g\text{ is conjugate to }g^{-1}.} \]