(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] Infinite Group Has Infinite Subgroups

The problem statement, all variables and given/known data

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, a_{1}, -a_{1}, a_{2}, -a_{2}, ...} 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.

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# Homework Help: Infinite Group Has Infinite Subgroups

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