= Solution
Suppose $\chi^\alpha\chi^\beta$ is irreducible. Since symmetric-group characters are real,
$$
1=\langle\chi^\alpha\chi^\beta,\chi^\alpha\chi^\beta\rangle
=\langle\chi^\alpha\chi^\alpha,\chi^\beta\chi^\beta\rangle.
$$
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 number>[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>.
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