# Infinite Group Has Infinite Subgroups

[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 <ai> ≠ <aj> if i ≠ j.

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

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

Related Calculus and Beyond Homework Help News on Phys.org
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.

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?

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.

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.

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.

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

Dick