Disjoint translates of invariant Salvetti cover members

Suppose $U(F,C)\cap gU(F,C)\mathrm{\ne }\mathrm{\varnothing }$ and put $p=(F,C)$. Naturality identifies the second member as $U(gF,gC)$. The chain criterion, or reflexivity if the indices coincide, implies that $p$ and $gp$ are comparable in $P(\mathcal{A})$.

Choose $h=g$ if $p\le gp$. Otherwise $gp\le p$, so applying the order automorphism $g^{-1}$ gives $p\le g^{-1}p$; in this case choose $h=g^{-1}$. Thus $p\le hp$. Because $G$ is finite, two powers $h^a,h^b$ with $a<b$ agree, giving $h^{b-a}=1$; put $m=b-a\ge 1$. Applying successive powers of the order automorphism $h$ gives $p\le hp\le \cdots \le h^{m}p=p$. Transitivity and antisymmetry therefore give $p=hp$, including when $m=1$. Since $h$ is $g$ or $g^{-1}$, it follows that $gp=p$.

The trivial pair-stabilizer conclusion now gives $g=1$. Consequently a nonidentity $g$ cannot satisfy $U(F,C)\cap gU(F,C)\mathrm{\ne }\mathrm{\varnothing }$, which proves the assertion.