Graphical regular representation - specific choice of generators From all I know the finite groups admitting a GRR are known completely. I currently try to use Godsil's results for some own ideas. That raises the following question: 
If $G$ is a given group that admits a GRR, is it possible to choose a generating set $S$ of $G$ such that the set $\left( \bigcup_{s \in S} \langle s \rangle \right) \backslash \{1\}$ is a generating set which provides a GRR of $G$? 
Or more generally: For a given group $G$ that admits a GRR, are there some subgroups $U_i \leq G$ such that the set $\left( \bigcup_i U_i \right) \backslash \{1\}$ provides a GRR?

I think the answer is no. Take $G$ to be the dihedral group of order 14, with presentation $\langle a,b|a^7=b^2=(ab)^2=1\rangle$ and suppose that $S$ is a generating set for $G$ of the type you require and such that $Cay(G,S)$ is a GRR.
If $S$ intersects $\langle a\rangle\cong C_7$ then $S$ must contain all the non-identity elements of $\langle a\rangle$. Taking graph complements if necessary, this implies that $G$ has a GRR with connection set containing only involutions.
Unless I made a mistake, I think this does not happen. (I checked, with some help from a computer.)
