# Why does Zorn's Lemma fail to produce a largest group?

Zorn's Lemma states that if every chain $$\mathcal{C}$$ in a partially ordered set $$\mathcal{S}$$ has an upper bound in $$\mathcal{S}$$, then there is at least one maximal element in $$\mathcal{S}$$.

Why can't we apply Zorn's Lemma in the following case?

Let $$\mathcal{S}$$ be the set of all groups. Define a partial order $$\preceq$$ as follows: for $$H, G \in \mathcal{S}$$, define $$H \prec G$$ if and only if $$H$$ is a subgroup of $$G$$. Then every chain $$\mathcal{C}=(G_{\alpha})_{\alpha \in A}$$ in $$\mathcal{S}$$ has an upper bound $$\cup_{\alpha \in A} G_{\alpha}$$ in $$\mathcal{S}$$. But certainly there is no maximal element in $$\mathcal{S}$$.

Could anyone tell me what is wrong with this argument?

• You can define partial orders on certain subclasses of groups. A fruitful one is the "largeness" ordering for finitely generated groups: $G\succeq H$ ("$G$ is larger than $H$") if there exists a finite-index subgroup $K\leq_fG$ such that $K\twoheadrightarrow H$. Exercise: Prove that free group $F_2$ is maximal here. Then a group $G$ is "large" or "large in the sense of Pride" (after Steve Pride, who first studied this ordering) if it is maximal in this ordering. For fruitfulness, here is a picture of Ian Agol announcing his proof of the virtual Haken conjecture. – user1729 Feb 7 '19 at 15:30
• Note that groups are really superfluous here: the same idea suggests that Zorn's lemma should yield a maximal set. This helps make it clearer that the class/set issue is absolutely fundamental. – Noah Schweber Feb 7 '19 at 15:34
• You might be able to apply the Hausdorff maximal principle instead, if you can generalize it to classes. – PyRulez Feb 7 '19 at 18:03
• I lost you at "set of all groups". – WillO Feb 8 '19 at 0:31

There is no set of all groups, but Zorn's Lemma can be phrased for partially ordered classes as well, but we need to require that any chain has an upper bound, also proper class chains. However in that case it is easy to define a proper class chain which has no upper bound.

Simply take for each ordinal $$\alpha$$ the free group generated by $$\alpha$$. There are natural injections given by the natural injections between two ordinals. However this chain does not have an upper bound, since an upper bound would be a group, which is a set, that embeds all ordinals. This is a contradiction to Hartogs theorem, of course.

(You can think about this in the following analog: an infinite chain of finite sets, or finite groups, will not have an upper bound which is also finite.)

• I guess you could also take for each $\alpha$ the set of all finite subsets of $\alpha$ with the symmetric difference operation, and then your natural injections are just inclusions. – bof Feb 7 '19 at 12:34
• True. That's also an option. The natural injections here are also just inclusions, though. – Asaf Karagila Feb 7 '19 at 13:46
• Doesn't that depend on what you mean by "the" free group generated by $\alpha$? I don't know any group theory, but I vaguely recall seeing free groups constructed as subdirect products. I guess you could also construct them as sets of words. Is that how you get the inclusions? Still seems more complicated than just using Boolean groups, if all you want is a proper class chain of groups. – bof Feb 7 '19 at 14:31
• Yes, I meant as a group of words. But I agree, your solution is simpler. – Asaf Karagila Feb 7 '19 at 14:46
• And not only is there no set of all groups, there's no set of groups which contains at least one group in each isomorphism class. – Daniel Schepler Feb 7 '19 at 22:36

Because there is no such thing as “the set of all groups”.