Hi

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,ifis abelian,thenit is cyclic.

So if is commutative, what about order ? Conlude