“…As such Q(D 3 ) = H(D 3 ), and that NHol(G) = QHol(G) ∼ = (S 3 × S 3 ) ⋊ C 2 where the C 2 component is that element conjugating λ(D 3 ) to ρ(D 3 ), so that π(Q(G)) is also this same subgroup of order 2. Indeed, if we embed D 3 as S 3 into S 6 then we have λ(S 3 ) = (1, 3)(2, 5) (4,6), (1,4,5)(2, 6, 3) ρ(S 3 ) = (1, 2)(3, 5) (4,6), (1,4,5)(2, 3, 6) QHol(S 3 ) = Hol(S 3 ) τ for any τ ∈ { (1,4), (1,5), (2,3), (2,6), (3,6), (4,5)} so that all π(Q(S 3 )) are isomorphic, which is unsurprising given that Q = H and QHol(S 3 ) = NHol(S 3 ) so that any π(Q(G)) would be isomorphic to T (S 3 ). For D n = x, t | x n = 1, t 2 = 1, xt = tx −1 in general, in [15], we have a complete enumeration of R(D n ) = R(D n , [D n ]).…”