I can't see how you deduce that is cyclic from what you've written before.
Lagrange's theorem (or one of his corollary): in a finite group, the order of an element divides the order of the group.
(For what follows, I assume you don't use Sylow's theorems.)
So the elements of have order 1, 5, 7 or 35. The only of order 1 is the identity element . Can the others all be of order 5? We want to show that no.
Let be an element of order . The subgroup generated by has five elements, and four different from . Using Lagrange's theorem, what are their order?
Now, do you agree that in a finite group of order , an element of order is a generator of the group?
So let be two elements of order 5 in , what if (i.e. they have a common element)?
Therefore, if all elements in have order 5, all cyclic subgroups have their intersections equal to , and the number of elements in is a multiple of 4, let's say where is a positive integer. Is possible? No? Then contradiction, and all elements can't have order 5.
Same thing with 7.
Finally, if some element has order 35 (i.e. is a generator of ), what about and ?
I guess that we've proved that contains at least an element of order 5 and an element of order 7.
Now, what we want to show is that given a of order 35, if is abelian, then it is cyclic.
So if is commutative, what about order ? Conlude