Let G be a finite group, and let π e (G) be the set of all element orders of G. In this short paper we prove that π e (B n (q)) = π e (C n (q)) for all odd q.Keywords finite group, element orders, Lie type simple group MSC 20D06, 20D60Let G be a finite group, and let π e (G) be the set of all element orders of G. Twenty years ago the author of this paper put forward the following conjecture [1]: Conjecture Let G be a group, and let M be a finite simple group. Then G ∼ = M if and only if (i) π e (G) = π e (M ), and (ii) |G| = |M |.For the above conjecture, Professor J. G. Thompson pointed that "This would certainly be a nice theorem" if we prove it. After this, a series of papers proved that this conjecture holds for all sporadic simple groups [1], the alternating groups A n (n 5) [2], the classical simple groups A n (q) [3], 2 A n (q) [4], D n (q) [5,6], 2 D n (q) [5], and all exceptional Lie type simple groups [7,8]. So, the last step of the proof of this conjecture is to prove that the conjecture holds for the simple groups B n (q) and C n (q). Since |B n (q)| = |C n (q)| and B n (q) ∼ = C n (q) for q odd, it is necessary to prove that π e (B n (q)) = π e (C n (q)) for all odd q.First we have the following lemma [9]:Lemma Let h, m be integers and h 2, m 3. If (h, m) = (2, 6), then there exists a prime r such that r | h m − 1, r h i − 1 for i < m. *