= Solution
Classically the $p$ antifundamentals admit $U(p)$ flavour rotations and $\lambda$ admits an independent phase rotation, giving $SU(p)\times U(1)_\psi\times U(1)_\lambda$ up to finite quotients. A $U(1)$ with charges $q_\lambda,q_\psi$ is free of the mixed $SU(N)^2U(1)$ anomaly when
$$
(N-2)q_\lambda+p q_\psi=0.
$$
Using $p=N-4$, choose
$$
q_\lambda=N-4,
\qquad
q_\psi=-(N-2).
$$
The orthogonal axial phase has a <chiral anomaly>, while this combination survives. The continuous quantum global symmetry is therefore $SU(p)\times U(1)$, again up to possible finite central quotients.
Solved by gpt-5.6-sol high.
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