A pointed metric-group category is modular exactly when its radical is trivial

The two hexagon equations in [F2] give $b(a+b,c)=b(a,c)b(b,c)$ after multiplying the formula for $r(a+b,c)$ by that for $r(c,a+b)$ and cancelling the three $\omega$ factors. Since $b(a,c)=b(c,a)$ by definition, it is multiplicative in the second variable as well. For $r\in \mathrm{rad}(b)$ nonzero, the rows indexed by $0$ and $r$ in [F1] coincide, since $b(r,h)=1=b(0,h)$ for every $h$. Thus $s$ is singular.

Because $G$ is finite and $b$ is a bicharacter, every $b(g,h)$ is a root of unity, so $\overline{b(g,h)}=b(g,h{)}^{-1}$. If $\mathrm{rad}(b)=\{0\}$, the characters $h\mapsto b(g,h)$ are pairwise distinct. For $g\mathrm{\ne }k$, $\chi (h)=b(g-k,h)$ is nontrivial; choose $t$ with $\chi (t)\mathrm{\ne }1$, and translating the finite sum gives ${\sum }_h\chi (h)=\chi (t){\sum }_h\chi (h)$, hence ${\sum }_h\chi (h)=0$. Thus distinct rows of $s$ are orthogonal and each row has squared norm $\mathrm{\mid }G\mathrm{\mid }$, so $s{s}^{†}=\mathrm{\mid }G\mathrm{\mid }I$. Therefore $s$ is invertible and the category is modular.