Proving: Infinite Group Has Infinite Subgroups

[bjnartowt]

Homework Statement

prove that an infinite group must have an infinite number of subgroups.

Homework Equations

The Attempt at a Solution

there are two cases to address regarding the order of the subgroups whose quantity is in question,
Case 1: those subgroups, themselves, are either finite or infinite
Case 2: those subgroups, themselves, are strictly finite.

suppose there was a member of G that had infinite order, which we will call x,
\exists x \in G{\rm{ }}\left| x \right| = \infty

Make iterations of that member x iterated,
\left\langle {{x^n}} \right\rangle

That constitutes an infinite subgroup. But there are more such infinite subgroups! There should be one such infinite subgroup every time we consider,
\exists \left\langle {{x^{{p_1}}}} \right\rangle

in which p1 is a prime number. Since there is an infinite number of primes, there is, correspondingly, an infinite number of \left\langle {{x^{{p_1}}}} \right\rangle that exist, and thus an infinite number of distinct subgroups of G, each of which are infinite in order!

This proof would be easy if only I could assume that an infinite group always has some element of infinite order. Then, you could chop up that infinite order into just-as-infinite subgroups (able to be iterated infinity times), and there would be one DISTINCT infinite subgroup per prime integer...of which there are infinite. Distinctness would fall from primeness.

 

[bjnartowt]

In fact, one author seems to imply that subtracting an infinite number of finite groups from an infinite group gives you a still-infinite group leftover after the aforementioned subtraction from it. I find that dubious!