= Solution
An unbroken <subgroup> $H$ must fix the <vacuum expectation value> itself. Thus a nonzero $H$-invariant <vacuum expectation value> can exist only if the <restriction of a representation> to $H$ contains a <trivial representation>. Applying the <singlet criterion for an unbroken subgroup> to <4/b/solution> gives
$$
\boxed{\begin{array}{c|cc}
R&\text{can preserve }H_2&\text{can preserve }H_3\\\hline
\mathbf4&\text{no}&\text{yes}\\
\mathbf6&\text{yes}&\text{no}\\
\mathbf{15}&\text{yes}&\text{yes}
\end{array}}
$$
The two “no” entries are forbidden for every nonzero <vacuum expectation value>. Each “yes” means that the indicated <subgroup> can remain unbroken, subject to the <scalar potential> choosing such a vacuum. It does \b[not] by itself prove that the indicated <subgroup> is the entire <stabilizer>.
For exact <spontaneous symmetry breaking> to the stated <subgroup>, the distinction matters. A complex $\mathbf4$ admits an exact $H_3$ <stabilizer>; a complex $\mathbf6$ admits an exact $H_2$ <stabilizer>. A single $\mathbf{15}$ preserving $H_2$ or $H_3$ necessarily preserves an additional continuous $U(1)$, as computed in <4/d/solution>, so neither exact endpoint is possible with that single field. Also, if the $\mathbf6$ <scalar field> is constrained to the <real structure of the SU4 exterior square>, its nonzero <stabilizer> is larger than $H_2$. The paper does not specify whether this reality constraint is imposed, so both cases are distinguished below.
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