Kronecker coefficients are nonnegative integers with full S_3 and conjugation symmetries

Claim (a): by [F1], $g(\lambda ,\mu ,\nu )$ is the multiplicity of an irreducible summand in a completely reducible finite-dimensional module, so it is a nonnegative integer.

Claim (b): by [F2], $g(\lambda ,\mu ,\nu )$ is the average over $\sigma \in S_n$ of the product ${\chi }^{\lambda }(\sigma ){\chi }^{\mu }(\sigma ){\chi }^{\nu }(\sigma )$. Permuting the three arguments permutes the three factors of each summand, and complex multiplication is commutative, so every summand and hence the average is unchanged; this gives all six orderings.

Claim (c): by [F3], ${\chi }^{{\lambda }^{\mathrm{\prime }}}(\sigma )=\mathrm{sgn}(\sigma ){\chi }^{\lambda }(\sigma )$, ${\chi }^{{\mu }^{\mathrm{\prime }}}(\sigma )=\mathrm{sgn}(\sigma ){\chi }^{\mu }(\sigma )$ and ${\chi }^{{\nu }^{\mathrm{\prime }}}(\sigma )=\mathrm{sgn}(\sigma ){\chi }^{\nu }(\sigma )$ for every $\sigma \in S_n$. Substituting the first two of these into the formula [F2] for $g({\lambda }^{\mathrm{\prime }},{\mu }^{\mathrm{\prime }},\nu )$ gives $\frac{1}{n!}{\sum }_{\sigma }\mathrm{sgn}(\sigma {)}^2{\chi }^{\lambda }(\sigma ){\chi }^{\mu }(\sigma ){\chi }^{\nu }(\sigma )=\frac{1}{n!}{\sum }_{\sigma }{\chi }^{\lambda }(\sigma ){\chi }^{\mu }(\sigma ){\chi }^{\nu }(\sigma )=g(\lambda ,\mu ,\nu )$, since $\mathrm{sgn}(\sigma {)}^2=1$ for every $\sigma$. The same computation with exactly two of the three substitutions gives $g({\lambda }^{\mathrm{\prime }},\mu ,{\nu }^{\mathrm{\prime }})=g(\lambda ,{\mu }^{\mathrm{\prime }},{\nu }^{\mathrm{\prime }})=g(\lambda ,\mu ,\nu )$.