= Solution
The breaking of the internal <global symmetry in field theory> $SU(n)\times SU(n)$ to $SU(n)$ removes $2(n^2-1)-(n^2-1)$ generators. The relativistic <Goldstone theorem> therefore requires
$$
\boxed{n^2-1\text{ massless Goldstone bosons}.}
$$
Other states fall into multiplets of the unbroken <special unitary group>, but their <masses> are not fixed by this counting. For the usual diagonal embedding $SU(n)_V\subset SU(n)_L\times SU(n)_R$, the <Goldstone bosons> form its adjoint multiplet.
If small explicit symmetry-breaking terms are added, the broken-generator directions generally acquire small curvature, producing light <pseudo-Goldstone bosons> rather than exactly massless states. Their squared <masses> tend to zero with those breaking parameters. Small breaking can also split multiplet degeneracies. These claims concern a small perturbation of the broken phase; approximate <symmetry> alone does not establish which <vacuum> is selected.
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