“…From [2, Lemma 2.5] the order of a soluble subgroup of N whose order is divisible by r should divide either m(q m − 1), or 2m 2 (m + 1)(q − 1)l with q = p, r = m + 1, m = 2 l . But in this last case, r = m + 1 is the only primitive prime divisor of such group with respect to the pair (q, m) and so, applying [21, 2.4 Proposition D] it should be (q, m) ∈ {(2, 4), (2,10), (2,12), (2,18), (3,4), (3,6), (5,6)}, but this contradicts the fact that m is a power of 2 and |π(G)| ≥ 5. Hence A ∩ N is a soluble subgroup of order dividing m(q m − 1).…”