Group theory help!

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.

Reply 1

Hopple

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

Reply 2

Gimothy

Have a look at Sylow's theorems.

Reply 3

Dirac Spinor

Original post by Gimothy

Have a look at Sylow's theorems.

wouldn't lagrange suffice?

Reply 4

nuodai

Suppose

G

is infinite, and split it into two cases: if

G

is finitely-generated then it contains an element of infinite order; and if

G

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

G

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

H \le G

then

\left| H \right|

divides

\left| G \right|

; it doesn't show that if

n

divides

\left| G \right|

then there is a subgroup of order

n

. (And rightly so -- this isn't true!)

Reply 5

Dirac Spinor

Original post by nuodai

No. That shows that if

H \le G

then

\left| H \right|

divides

\left| G \right|

; it doesn't show that if

n

divides

\left| G \right|

then there is a subgroup of order

n

. (And rightly so -- this isn't true!)

ooops, right you are :biggrin:

Reply 6

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.

(edited 12 years ago)

Thanks TSR.

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.

Reply 9

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.

Reply 10

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.

Reply 11

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?

Reply 12

DFranklin

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

Reply 13

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.

Reply 14

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.

Reply 15

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?