Let G be a group of order 35 = 5 * 7.
So, G has p-Sylow subgroups of orders p = 5, 7; let m and n (respectively) be the number of these subgroups.
Then, we have
(i) m|5 and m = 1 (mod 7) ==> m = 1.
(ii) n|7 and n = 1 (mod 5) ==> n = 1.
Hence, both subgroups are normal in G.
Being of prime order, these are cyclic of orders 5 and 7, which has trivial intersection.
Therefore, G ≅ Z5 x Z7 ≅ Z35, which is cyclic.
I hope this helps!