 Solution

Suppose ${\chi }^{\alpha }{\chi }^{\beta }$ is irreducible. Since symmetric-group characters are real,Both self-products contain the trivial character once. They can therefore have no other common irreducible constituent. By part i, the standard character occurs in the two self-products with multiplicities $|{\alpha }^{-}|-1$ and $|{\beta }^{-}|-1$. Hence one of these numbers is zero; say $|{\alpha }^{-}|=1$.

A partition has exactly one removable node precisely when all its nonzero rows have equal length, so $\alpha =(a^b)$ is rectangular. Since $ab=n$ and $n$ is prime, either $a=1$ or $b=1$. Thus $\alpha =(1^n)$ or $(n)$. The same argument applies with $\alpha$ and $\beta$ interchanged, proving the prime-degree irreducible Kronecker product criterion for a symmetric group.