For a given group $(G,+)$, $+$ the usual addition and $G$ a set where every finite subset has a maximum then every subgroup of $G$ is infinite

Question

I want to prove that given $(G,+)$ a group whith $+$ the usual addition and $G$ a group such that every finite subset has a well defined maximum then there is not any finite subsets in $G$

My proof goes the following way:

Lets suppose that there is $S$ a subgroup of $G$ such that $\#S=n\in \mathbb{N}$. Then, by taking $m:=max(S)$ we get that $m+m \notin S$ therefore $S$ isn't a subgroup and we get to an absurd.

This way we are prooving that for every subgroup $S$, $\#S=\infty$

This can be applyed for say for instance $(Q, +)$

Being this such an abstract proof I'd like someone to check it and let me know wether it's correct. In that case I'd be interested to know if the conditions here could be weakened somehow.

Answer 1

As many commenters have pointed out, your question, as written, doesn't make much sense. However it's easy to guess what kind of question you wanted to ask, and your intuition and proof will basically work in this new setting.

First, the repaired question:

Given an (abelian) linearly ordered group $(G,\leq)$, must every nontrivial subgroup be infinite?

Here an (abelian) group $(G,+,-,0)$ is called "linearly ordered" if it comes equipped with a linear order $\leq$ which is compatible with addition in the sense that $a \leq b$ implies $c+a \leq c+b$.

Originally you ask about groups with an ordering so that "every finite subset has a maximum". This condition is exactly the condition that $\leq$ be a linear order, so this really is a direct rephrasing of your question! Moreover, the proof you attempt to give is only making use of the order and group structure, plus implicitly the fact that they're "compatible" in the sense outlined above. So it should go through perfectly well in this setting! There's a few tweaks I had to make, but you had the exact right idea!

Indeed, let $S$ be a finite subset of $G$ containing a nonzero element. Then $S$ cannot be a subgroup. Indeed, let $s^*$ be the maximal nonzero element of $S$. If $0 \leq s^*$, then by compatibility we see $s^* + 0 \leq s^* + s^*$ so that $s^* + s^*$ cannot be in $S$, as it is larger than the maximal element. If instead $s^* \leq 0$, then $-s^* \geq 0$ cannot be in $S$, as again it is larger than the maximal element. Either way, $S$ is not a subgroup, and we are done.

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Now, briefly, why did I put "abelian" in parentheses above? I wanted to stick to your notation for $G$, which was additive, and culturally that usually means that $G$ should be abelian. But notice that we didn't actually use the abelian assumption anywhere! In the nonabelian setting, we usually only require the compatibility condition to be true when we multiply on the left. In this case $G$ is called left orderable, and the same proof shows that left orderable groups cannot have finite subgroups!

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I hope this helps ^_^