Is there a finite group $G$, an element $c$ of order 2 in $G$, and an irreducible 2-dimensional complex representation $\rho$ of $G$ such that all the following are true:
1) $\rho(c)$ has trace zero
2) There is an index 2 subgroup $H$ of $G$ containing $c$, and a 1-dimensional representation $\psi$ of $H$, such that $\rho\cong Ind(\psi)$, the representation of $G$ induced by $\psi$.
3) If $K$ is any index 2 subgroup of $G$ and $\chi$ any 1-dimensional representation of $K$ such that $\rho\cong Ind(\chi)$, then $c\in K$.
Here are some thoughts, but nothing really concrete.
If a 2-dimensional representation is induced from a 1-dimensional representation of an index 2 subgroup, then one checks easily that the trace of the 2-d representation vanishes off the index 2 subgroup. So assumption (2) above implies that the trace of $\rho$ will vanish off $H$, and (1) says "the trace of $\rho$ vanishes on $c$ too", but then (3) says "...but that is not because $\rho$ is induced from an index 2 subgroup not containing $c$".
Note for people wondering about the difference between (2) and (3): it is possible that a 2-dimensional representation of a finite group can be induced from 1-dimensional representations of several distinct subgroups (consider for example the faithful 2-dimensional representation of the quaternion group of size 8).
[ Background: this question came up because the question arose about whether a certain kind of theta series could exist; the theta series, if it existed, would give rise to a certain kind of modular form; if the theta series existed, and my understanding is correct, then $G$ would be the Galois group of a certain extension of number fields, $\rho$ would be the representation attached to the form and $c$ would be some complex conjugation. The details are a bit technical, but basically if one could rule out $\rho$ above using purely representation-theoretic methods then one could rule out the existence of the theta series, and the question above seems to me to be more tractible. If however $\rho$ does exist then the question I'm actually interested in would still be open because the implication only goes one way.]