Infinite Group Has Infinite Subgroups

[e(ho0n3]

[SOLVED] Infinite Group Has Infinite Subgroups

Homework Statement
Prove that an infinite group must have an infinite number of subgroups.

The attempt at a solution
There are two infinite groups that I can think of: the additive group of integers Z and the multiplicative group of positive reals R+. Except for 0, all elements of Z have infinite order and <n>, n ≥ 0, is a unique subgroup of Z. The same seems to hold for R+. Hmm...

Let G = {e, a₁, -a₁, a₂, -a₂, ...} be an infinite additive group. I'm led to believe that <a_i> ≠ <a_j> if i ≠ j.

The following is a plausible proof by contradiction: Suppose <a_i> = <a_j>. That means <a_i> is a subset of <a_j>, so there is a positive integer m > 1 such that ma_j = a_i, and <a_j> is a subset of <a_i> so there is a positive integer n > 1 such that na_i = a_j. This implies that mna_i = ma_j = a_i and since mn > 1, <a_i> is finite which means that a_i has finite order. So all I have to prove now is that a_i has infinite order.

This is where I'm stuck. I need a little push.

 

[zhentil]

If there exists x in G such that x has infinite order, consider the group generated by x^2, x^3, x^4, ... Containment one way is trivial, so if the groups aren't distinct, I'm sure you can find a contradiction to the fact that x has infinite order.

If |x| is finite for all x in G, pick x1, x2, x3, ... such that x2 is not contained in <x1>, etc. These are an infinite number of subgroups, and are obviously distinct since otherwise |G| is finite.

 

[e(ho0n3]

If there exists x in G such that x has infinite order, consider the group generated by x^2, x^3, x^4, ... Containment one way is trivial, so if the groups aren't distinct, I'm sure you can find a contradiction to the fact that x has infinite order.

Right. This is what I did in my first post, essentially.

If |x| is finite for all x in G, pick x1, x2, x3, ... such that x2 is not contained in <x1>, etc. These are an infinite number of subgroups, and are obviously distinct since otherwise |G| is finite.

So I start by picking x₁ in G. Then I find an x₂ in G that is not in <x₁>. Then I find an x₃ in G that is not in <x₂> and <x₁>. Etc. And there is an infinite number of these. Sure they are distinct but how does that make |G| finite?

 

[zhentil]

It doesn't make |G| finite. That's the point. You've found an infinite number of distinct subgroups. If the process terminated at some point, that would make |G| finite, a contradiction.

 

[e(ho0n3]

If the process terminated at some point, that would make |G| finite, a contradiction.

I messed up. I meant to ask why |G| is finite if the process terminates.

 

[zhentil]

Because you can count the elements. Say it stops after n steps. Then you can't find an element of G not in <xi> for some i. But since |<xi>| is finite for each i, you have a finite sum of finite numbers, and you're done.

 

[e(ho0n3]

Right! I forgot that |<x_i>| = |x_i| which is finite. Thanks a lot.

 

[Dick]

This is a little late, but you might also notice that your first attempt was doomed to failure. To say <ai>=<aj> iff ai=aj only true in some groups (such as both of your examples). Consider Z_p for p prime. All non-identity elements generate the whole group. You might say that's because it's finite, but then consider Z_p x Z.