G a group of order 35, H a normal subgroup of order 7. Prove that if g in G has order 7, then g is in H.
What I'm doing is invoking Sylow's third theorem to show that there exist a single subgroup of order 5 and a single subgroup of order 7. And then the direct product of these has order 35, and then somehow... this means G has an element of order 35, which would mean G is cyclic, and if G is cyclic, each of its subgroups are cyclic. So H has an element of order 7 and if there's an element g in G of order 7, then it generates the subgroup H?
This is my train of logic? I could use some help on the somehow... part, or if that's even a valid way of proving that a group of order 35 is cyclic. Please help steer me back on track.