Thread: Group: order 35

Having troubles with this, i know that 35 = 7x5, which are both prime and subsequently 5 doesn't divide 6, so G is cyclic.

But I don't understand how to show there are elements of order 5 and 7 exactly, and that it's commutative, i.e abelian. Thanks in advance.

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, if is abelian, then it is cyclic.

So if is commutative, what about order ? Conlude

This is a proof that uses a Sylow theorem.

Lemma 1. If there is only one Sylow p-subgroup P in G, then P is normal in G.

By Sylow's theorem, G has subgroups of order 5 and order 7.

If the order of G is pq, then the number of Sylow q-subgroup is kq+1 and divides pq.
We see that there exists only one Sylow 5-subgroup and one Sylow 7-subgroup in G. Thus, both are normal in G by lemma 1.

Since both <5> and <7> are normal in G and intersects only at e in G, G = <5> <7> . Thus G is abelian. Since (see finitely generated abelian group), G is cyclic.