“…Now these include the eight pairs (3,15), (3,24), (3,42), (4,20), (6,12), (8,24), (9,18) and (12,12), with |G| = 2µ(k, ℓ)p = 20p, 16p, 14p, 10p, 8p, 6p, 6p and 6p, respectively, but as p ≥ 5, none of the listed group orders is divisible by the first entry k of the corresponding pair (k, ℓ), contradicting the fact that one generator of G has order k. Hence we can eliminate all of them from consideration. Also we can eliminate two further pairs, namely (7,42) and (5,20): in the former case, |G| = 2µ(k, ℓ)p = 6p and is divisible by ℓ = 42, and so p = 7, but then G ∼ = C 42 which is clearly not (7, 42, 2)-generated, while in the latter case |G| = 2µ(k, ℓ)p = 8p and is divisible by ℓ = 20, so p = 5, and then |G| = 40 and so G has a normal Sylow 5-subgroup P , but then the quotient G/P has order 8 and so is clearly not (1, 4, 2)-generated.…”