Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2010 / B / 6 / 1

There is more than one solution: the trivial representation $\lambda =(6)$ has character $1$ everywhere and the sign representation $\lambda =(1^6)$ has character $\pm 1$ everywhere. ThusIn fact they are the only ones. The full-cycle formula in the preceding discussion forces any candidate to be a hook $(6-r,1^r)$. For $r=1$, the standard character is the number of fixed points minus one, so it vanishes on cycle type $(5,1)$; its sign twist $r=4$ does too. For $r=2$, the hook representation is the exterior square of the standard representation and has character $\frac{1}{2}(\chi (g{)}^2-\chi (g^2))$. At cycle type $(4,1,1)$ both standard-character values are one, so this character is zero. Its sign twist $r=3$ also vanishes there. Only $r=0,5$ remain.