[Blutooth]

Let G be a nontrivial group (remember that the trivial group is the group with only one element, so a
nontrivial group is a group with at least 2 elements). Suppose that the only subgroups of G are G and
{1}. Prove that G is cyclic and finite, and prove that the number of elements in G is a prime number.

I can do the first part but not the second in bold, help would be appreciated.

[Hopple]

If it isn't prime, can you show that it has some subgroups other than those stated?

[Gimothy]

Have a look at Sylow's theorems.

[ben-smith]

(Original post by Gimothy)
Have a look at Sylow's theorems.

wouldn't lagrange suffice?

[nuodai]

Suppose is infinite, and split it into two cases: if is finitely-generated then it contains an element of infinite order; and if is infinitely-generated then it contains infinitely many generators (duh). In each of these cases you can find a proper subgroup by choosing every second power or every second generator, say. I'll let you fill in the details.

Now suppose is finite. If it is generated by more than on element, can you find a proper subgroup? What about if it is generated by precisely one element which has composite order?

(Original post by ben-smith)
wouldn't lagrange suffice?

No. That shows that if then divides ; it doesn't show that if divides then there is a subgroup of order . (And rightly so -- this isn't true!)

[ben-smith]

(Original post by nuodai)
No. That shows that if then divides ; it doesn't show that if divides then there is a subgroup of order . (And rightly so -- this isn't true!)

ooops, right you are

[Blutooth]

(Original post by Hopple)
If it isn't prime, can you show that it has some subgroups other than those stated?

I think I might be on the right tracks if I start by assuming if

G=<g>, then write the identity element as g^(a) for some integer a and then do some more stuff. Will get back to you soon.

Ok, I think that g^n=e must be the case. And because of this if n=ab a,b>1, we can take any (g^a), (g^2a),...,(g^ba) will form a subgroup. Think need to prove this a bit more rigorously but I'm starting to see it. If I'm on the wrong tracks please tell me.

[Blutooth]

Thanks TSR.

[Gimothy]

(Original post by ben-smith)
wouldn't lagrange suffice?

Nope, Sylow gives a partial converse to Lagrange, as a poster above has detailed I think.

[DFranklin]

(Original post by Gimothy)
Nope, Sylow gives a partial converse to Lagrange, as a poster above has detailed I think.

But it's pretty much overkill here, I think.

[SsEe]

(Original post by Gimothy)
Nope, Sylow gives a partial converse to Lagrange, as a poster above has detailed I think.

Does Sylow obviously cover the case where the order of G is p^n with p prime and n > 1 ? I'd use Cauchy's theorem instead. It's lighter and covers this case cleanly.

[Totally Tom]

(Original post by SsEe)
Does Sylow obviously cover the case where the order of G is p^n with p prime and n > 1 ? I'd use Cauchy's theorem instead. It's lighter and covers this case cleanly.

Isn't Cauchy's theorem just a trivial consequence of Sylow's theorems?

[DFranklin]

Am I being stupid, or did Nuodai answer this completely without using Sylow/Cauchy/etc?

[around]

Every subgroup of a cyclic group is normal (because a cyclic group is abelian) and so appeal to the classification theorem for finite simple groups.

[Totally Tom]

(Original post by DFranklin)
Am I being stupid, or did Nuodai answer this completely without using Sylow/Cauchy/etc?

Nuodai is workin in ZFC, pretty sure the OP wanted a solution working within ZF.

[DFranklin]

(a) I'm not sure why you think that, and (b) surely ZF v.s. ZFC only arises in the case G is infinite and since Cauchy/Sylow are for finite groups they can't help anyhow?

[Totally Tom]

Srs bsnizz.