- #1

bjnartowt

- 284

- 3

## 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,

[tex]\exists x \in G{\rm{ }}\left| x \right| = \infty [/tex]

Make iterations of that member x iterated,

[tex]\left\langle {{x^n}} \right\rangle [/tex]

That constitutes an infinite subgroup. But there are more such infinite subgroups! There should be one such infinite subgroup every time we consider,

[tex]\exists \left\langle {{x^{{p_1}}}} \right\rangle [/tex]

in which p1 is a prime number. Since there is an infinite number of primes, there is, correspondingly, an infinite number of [tex] \left\langle {{x^{{p_1}}}} \right\rangle [/tex] 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.