# Zorn's lemma converse? (Context: Maximal proper subgroups)

So, in my qual prep class a pretty simple question popped up: "Prove that for any nontrivial finite group there exists a maximal proper subgroup."

So of course, my natural inclination was to wonder if we needed the finite assumption, as it seemed like the standard Zorn's lemma argument would apply (Take the union of the a partially ordered chain). After some discussion, the problem came up with how do we know such a union is actually a proper subgroup?

After some thought, I'm pretty sure I came up with a counterexample: Take the group generated by rational numbers of the form $\frac 1 {2^n}, \forall n\in \mathbb N$ under addition. So any fraction in which the denominator is a power of 2 is our base group. Then we could have a chain where each group is generated by $\frac 1 {2^n}$, and we see that they are partially ordered by inclusion, each one is a proper subgroup, but the union of the chain is the whole group, thus is not a proper upper bound.

So far, so good. My question came in...is this sufficient to prove that a maximal element does not exist? I.e., does the converse of Zorn's lemma hold?

• Could you be more specific to how you would phrase the converse to Zorn? Anyway, the rationals has no maximal subgroups (good exercise). – Tobias Kildetoft May 20 '15 at 9:51
• Your (counter)example $\mathbb Z[\frac12]$ shows that Zorn's lemma is not applicable for this problem for arbitrary groups. – Hagen von Eitzen May 20 '15 at 9:51
• The converse I'd be looking for would be is this an if and only if statement? I.e., if there exists such a chain with no upper bound, does that exclude the possibility of a maximal element? – Alan May 20 '15 at 9:53
• ...on second thought, forget this. It's obvious that's not true. Wow. Okay, sleep dep :) – Alan May 20 '15 at 9:54
• What is "the converse of Zorn's lemma" here? Finiteness is essential, as you see, without appealing to Zorn's lemma at all. The axiom of choice has nothing to do with this proof to begin with, since the general statement is not true in $\sf ZFC$ anyway. – Asaf Karagila May 20 '15 at 10:28