# Can i assert that a gruop of infinite order has an element of infinite order, always?

#### bjnartowt

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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.

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#### bjnartowt

Re: can i assert that a gruop of infinite order has an element of infinite order, alw

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!

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