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**[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, 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.