 Solution

One left-handed Standard Model family requires the SU(5) representation content $10\oplus \overline{5}$. No single irreducible representation in solution contains both. At the level of containing the required fields, the smallest choice isThe additional SU(5) multiplets are a $5$ and a trivial representation. Another minimal choice using only two of the listed irreducible representations isIts additional multiplets form the conjugate representation of a family. More generally, for multiplicities $(m_6,m_{\bar{6}},m_{15},m_{\overline{15}})$, branching of a representation givesThus containing a family requires $m_{15}\ge 1$ and $m_{\bar{6}}+m_{\overline{15}}\ge 1$.

For a consistent chiral grand unified theory, containment alone is insufficient. Normalize the cubic gauge anomaly coefficient by $A(6)=1$. For the special unitary group $SU(N)$, $A({\Lambda }^2\mathbf{N})=N-4$, so for SU(6) it is $2$. To verify this directly, take a traceless diagonal generator with entries $x_i$: its exterior square has entries $x_i+x_j$, andThe conjugate representation reverses this cubic gauge anomaly. The combination $15\oplus \overline{6}$ has coefficient $2-1=1$, so cannot alone be a quantum-consistent chiral SU(6) gauge theory. The minimal anomaly-free choice with one net chiral gauge spectrum of a family is insteadThe additional nonsinglet fields form a vectorlike gauge spectrum that can acquire an SU(5)-invariant mass, and the two gauge singlets may serve as sterile neutrinos. More generally, precisely one net chiral gauge spectrum of a family with vectorlike extra nonsinglets requires $m_{15}-m_{\overline{15}}=1$ and $m_{\bar{6}}-m_6=2$; these conditions also cancel the cubic SU(6) gauge anomaly. The $15\oplus \overline{15}$ choice above is anomaly-free but has zero net chirality.