Let R be a semiprime ring and f be an endomorphism on R. If f is a strong commutativity preserving (simply, scp) map on a non-zero ideal U of R, then f is commuting on U. A ring R is said to be prime if aRb = 0 implies that either a = 0 or b = 0, and semiprime if aRa = 0 implies that a = 0 where a; b 2 R: A prime ring is obviously semiprime. If R is a ring and S µ R, a mapping f : R ! R is called strong commutativity preserving (simply, scp) on S if (x;y) = (f(x);f(y)) for all x;y 2 S; and commuting on S if (f(x);x) = 0 for all x 2 S. For recent references on the commutativity in prime and semiprime rings, see (1) and (3); and for scp maps see (2) and (4). To prove the main result we need the following lemma which is of independent interest and can be used for further investigation. Lemma 1. If R is a semiprime ring and f is an endomorphism on R which is scp on a non-zero right ideal U, then for all x 2 U, f(x) i x commutes with (U;U). Proof. For all x;y 2 U, we have (x;xy) = (f(x);f(xy)). This implies that x(x;y) = f(x)(x;y) and so (f(x) i x)(x;y) = 0: From (x;yx) = (f (x);f (yx)) we can similarly show (x;y)(f(x) i x) = 0: For all x;y 2 R, replacing y by yr, we get (f(x) i x)y(x;r) = 0: This implies that (f(x) i x)U(x;r) = 0 and so (f(x) i x)UR(x;r) = 0: