“…Let m 2 = 2, m 3 = 1. Then by [3], there exist x, y, z ∈ G such that b 1 = (x, y), b 2 = (x, z), c 1 = (x, y, y), G ′ = b 1 , b 2 , c 1 , d 1 , γ 3 (G) = c 1 and γ 4 (G) = d 1 Therefore by Lemma 2.1, (b 1 − 1) 2 (b 2 − 1) 2 (c 1 − 1) 2 (d 1 − 1) 2 ∈ KG[14] = 0, a contradiction. Suppose m 2 = m 3 = 1, then as in Lemma 2.5, G ′ is abelian and |G ′ | = 3 3 .…”