Solution
= Solution
Take $g$ to be an $n$-cycle. For $n\geq2$, the <Murnaghan–Nakayama rule> gives $\chi^{(n)}(g)=1$ and $\chi^{(n-1,1)}(g)=-1$; every remaining two-row diagram contains a $2\times2$ square and is not a rim hook, so its value at $g$ is zero. Therefore
$$
F(g)=1-(-1)=2\ne0.
$$
The exhibited cycle has length $k=n$, so $k\equiv n\pmod2$. For $n=1$, the sole character has value one at the identity and the same conclusion holds with $k=1$.