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

By Scarlet

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?

#### @Asaf Karagila 2019-02-07 11:49:13

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.)

#### @bof 2019-02-07 12:34:47

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.

#### @Asaf Karagila 2019-02-07 13:46:05

True. That's also an option. The natural injections here are also just inclusions, though.

#### @bof 2019-02-07 14:31:17

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.

#### @Asaf Karagila 2019-02-07 14:46:45

Yes, I meant as a group of words. But I agree, your solution is simpler.

#### @PyRulez 2019-02-07 18:05:23

Does the Hausdorff maximal principle work with classes? If so, you could create create an infinite chain of groups, and then take their union of get a class sized group.

#### @Asaf Karagila 2019-02-07 18:10:29

@PyRulez: Well, what exactly do you mean?

#### @PyRulez 2019-02-07 18:14:00

@AsafKaragila Like, it would be achieving a slightly different goal, but you could use the Hausdorff maximal principle to get a maximal total order of groups, and then take their union to make a Group that is maximal.

#### @Asaf Karagila 2019-02-07 22:12:17

@PyRulez: You're implicitly assuming that such maximal chain is going to be a set. It won't be. No set of groups can be maximal. There's always another group, larger than all of the ones you have.

#### @PyRulez 2019-02-07 22:12:50

@AsafKaragila I meant a class sized group.

#### @Asaf Karagila 2019-02-07 22:15:59

Well, for a class-sized group, that's certainly possible to have a maximal one. Simply endow the entire universe with a group structure. (e.g. using symmetric difference). Clearly there is no larger group, since you've used up all the sets.

#### @Daniel Schepler 2019-02-07 22:36:01

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.

#### @José Carlos Santos 2019-02-07 11:34:57

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

#### @jasmine 2019-02-14 12:00:00

+1 jose carlos sir