= Solution
Let $V_i=\partial V/\partial\phi_i$ and $M^2_{ij}=\partial_i\partial_jV|_{\phi_0}$ be the <scalar mass matrix>, which is the <Hessian matrix> of the <scalar potential> at the vacuum. Invariance under the infinitesimal <Lie group action> gives
$$
V_i(\phi)(it^a\phi)_i=0.
$$
Differentiate with respect to $\phi_j$ and evaluate at the <vacuum expectation value> $\phi_0$. Since $V_i(\phi_0)=0$ at a minimum,
$$
M^2_{ji}(it^a\phi_0)_i=0.
$$
Thus every tangent vector $it^a\phi_0$ generated by a broken <Lie algebra generator> is a <zero eigenvalue> eigenvector of the <scalar mass matrix>. The unbroken generators are precisely those in the stabilizer Lie algebra $\mathfrak h$ and give the zero tangent vector; independent broken directions span $\mathfrak g/\mathfrak h$. Hence <Goldstone theorem> gives
$$
\boxed{\dim G-\dim H}
$$
massless scalar modes at the classical level.
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