# A “mother of all groups”? What kind of structures have “mother of all”s?

For ordered fields, we have a “mother of all ordered fields”, the surreal numbers $\mathbf{No}$, a proper-class “field” which includes (an isomorphic copy of) every other ordered field as a subfield. So, I wondered: does a similar mother object exist for other kinds of object, like groups? That is, is there some “meta-group” that's a proper-class “group” which every other group is a subgroup? If so, how could it be constructed?

My attempt at a construction was this. It is based on Cayley's theorem, that every group is isomorphic to some group of permutations. In particular, the symmetric groups on $n$ letters, $S_n$, can be considered as “mother groups” for all finite groups of order $n$ or less: every such group is embedded isomorphically as a subgroup of $S_n$. It is also possible to see how this holds for infinite groups as well: there are groups $\mathrm{Per}_{\kappa}$ for infinite cardinals $\kappa$ of all permutations (bijective functions) of a set of cardinality $\kappa$ that act as mother groups for all groups of cardinality $\kappa$ or less.

So what I imagine was the following idea. We start with a class of sets such that:

1. they are ordered by inclusion,
2. every cardinality is represented exactly once

We here use the initial ordinals of cardinals, interpreted as sets in the usual way (we assume the axiom of choice here). Then we consider building up on each initial ordinal $I_{\alpha}$ (which is indexed with ordinals $\alpha$ such that as we run up from 0 through to $\omega$, but not reaching $\omega$, $I_{\alpha}$ is the finite ordinal $\alpha$, then $I_{\omega}$ is $\omega$, the initial ordinal of cardinal $\aleph_0$, then $I_{\omega+1}$ is $\omega_1$, the initial ordinal of cardinal $\aleph_1$, etc.) its corresponding group of permutations, $\mathrm{Sym}(I_\alpha)$. We now imagine the class-sized mega-union of all these $\mathrm{Sym}(I_\alpha)$, i.e.

$$\mathbf{SYM} = \bigcup_{\alpha \in \mathbf{On}} \mathrm{Sym}(I_\alpha)$$.

Next, we proceed to define a composition of permutations in $\mathbf{SYM}$. Let $p$ and $q$ be two such permutations. If $\mathrm{dom}(p) = \mathrm{dom}(q)$, then their composition $p \diamond q = p \circ q$ – the usual composition. However, if $\mathrm{dom}(p) \ne \mathrm{dom}(q)$, then we have to first extend one or the other permutation. Define, for $\mathrm{dom}(p) < \mathrm{dom}(q)$,

$$p^q: \mathrm{dom}(q) → \mathrm{dom}(q)$$

$$p^q(a) = \begin{cases}p(a),&\ \mathrm{if}\ a \in \mathrm{dom}(p)\\ a,&\ \mathrm{otherwise}\end{cases}$$

.

Then, if $\mathrm{dom}(p) < \mathrm{dom}(q)$, $p \diamond q = p^q \circ q$, and if $\mathrm{dom}(q) < \mathrm{dom}(p)$, $p \diamond q = p \circ q^p$. We then define an equivalence relation ~ on permutations in $\mathbf{SYM}$ such that two are equivalent if one can be extended to the other in the fashion above. Then an operation between equivalence classes of composition can be defined by taking two representative permutations and composing. Of course, this runs into a difficulty since we cannot collect these equivalence classes together as they themselves are proper classes. But we can take the representative defined on the lowest possible $I_\alpha$. Denote this lowest representative of a permutation $p$ by $\mathrm{low}(p)$. Now we should have a “mother of all groups”, given by the class of all $\mathrm{low}(p)$ for every $p \in \mathbf{SYM}$, with composition defined by taking the low of the composition as defined before.

My questions are: does the above construction make any sense? If not, where's the flaw? If so, is there a way to get around the obvious use of the axiom of choice that I mentioned? One thing I notice about rejecting choice is that then the cardinals are not necessarily totally-ordered any more, so then however we'd go about choosing representatives, we'd run into the problem where we could not “nest” them together and so could not extend the permutations so as to be able to perform the composition. Does this mean the existence of the mother group depends on the axiom of choice? Also, what about “mother objects” of other types? I suspect that in a manner analogous to the above, we can construct a “mother of all rings” by the corresponding Cayley's theorem analogue for rings (rings are isomorphic to rings of endomorphisms of abelian groups). So this makes me wonder: what kind of conditions are required for some kind of structure to have a “mother structure” of proper-class size? Does any structure have one or just some?

Another thing I notice here is that this “mother group” seems not to include all proper-class “super” groups, only “normal”, i.e. set, groups, whereas, I think, No includes all proper-class “super” ordered fields as well. Which makes me wonder: what kind of criteria are needed to ensure the existence of a “true mother” proper-class version of a structure that includes all other such structures including other proper-class ones as subsets/classes? What do ordered fields have that groups don't, and what else besides ordered fields share this property?

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At least for finite groups there is this: en.wikipedia.org/wiki/Hall%27s_universal_group. –  Michael Albanese Apr 1 '13 at 8:10
Perhaps you're thinking of a generalised Fraïssé limit: en.wikipedia.org/wiki/Age_%28model_theory%29 –  Zhen Lin Apr 1 '13 at 10:54
50 years ago, there was a sort of riddle going around: "What do you call the functor from which all other functors are derived?" "The mother functor." –  Carl Weisman Apr 1 '13 at 12:17
I don't know if this refers to the newsletter? motherfunctor.org –  Martin Brandenburg Apr 5 '13 at 23:06

The surreal numbers exhibit much stronger universal properties than you have mentioned, for they also exhibit very strong homogeneity and saturation properties. For example, every automorphism of a set-sized elementary substructure of the surreals extends to an automorphism of the entire surreal numbers, and every set-sized type over the surreals that is consistent with the theory of the surreal numbers is realized in the surreal numbers. That is, any first-order property that could be true about an object as it relates to some surreal numbers, which is consistent with the theory of the surreal numbers, is already true about some surreal number.

For any complete first-order theory $T$, one can consider the concept of a monster model of the theory. This is a model $\mathcal{M}$ of $T$, such that first, every other set-sized model of $T$ embeds as an elementary substructure of $\mathcal{M}$--not merely as a substructure, but as a substructure in which the truth of any first-order statement has the same truth value in the substructure as in big model--and second, such that every automorphism of a set-sized substructure of $\mathcal{M}$ extends to an automorphism of $\mathcal{M}$.

One may also use approximations to these proper class monster models by considering extremely large set-sized models, of some size $\kappa$, such that the embedding and homogeneity properties hold with respect to substructures of size less than $\kappa$.

Every consistent complete first order theory $T$ has such monster models, and they are used pervasively in model theory. Model theorists find it convenient, when considering types over and extensions of a fixed model, to work inside a fixed monster model, considering only extensions that arise as submodels of the fixed monster model.

Your topic also has a strong affinity with the concept of the Fraïssé limit of a collection of finitely-generated (or $\kappa$-generated) structures, which one aims to be the age of the limit structure, the collection of finitely-generated ($\kappa$-generated) substructures of the limit structure. Fraïssé limits are often built to exhibit the same saturation and homogeneity properties of the surreal numbers. As a linear order, the surreal numbers are the Fraïssé limit of the collection of all set-sized linear orders. And I believe that one can also put the field structure in here.

Update. When one has the global axiom of choice, the set-homogeneous property of the generalized Fraïssé limit allows one to establish universality for proper class structures. Basically, using global AC one realizes a given proper class structure as a union of a tower of set structures, and gradually maps them into the homogeneous structure. The homogeneity property is exactly what you need to keep extending the embedding, and so one gets the whole proper class structure mapping in. This kind of argument, I believe, shows that what you want to consider is homogeneity rather than merely the universal property itself. (This argument is an analogue of the idea that when one has homogeneity for countable substructures, then one gets universality for structures of size $\aleph_1$.)

Regarding your specific construction, here is a simpler way to undertake the same idea, which avoids the need for the equivalence relation: Let $G$ be the proper class of all fixed-point-free permutations of a set. This class supports a natural group operation, which is to compose them, regarding elements outside the domain as fixed by the permutation, and then cast out any newly-created fixed points. The identity element of $G$ is the empty function, which is really a stand-in for the identity function on the universal class.

It is clear that every group finds an isomorphic copy inside $G$, without using the axiom of choice, since every group is naturally isomorphic to a group of permutations, and these are naturally embedded into $G$, simply by casting out fixed-points. This does not use the axiom of choice.

The class group $G$ is a natural presentation of the set-support symmetric group $\text{Sym}_{\text{set}}(V)$ of the set-theoretic universe $V$, the class of all permutations of $V$ having set support. Any such permutation of $V$ is represented in $G$ by restricting to the non-fixed-points.

Note finally that in the case of the surreal numbers, one doesn't need the axiom of choice in order to construct the surreal numbers, and one can get many universal properties for well-orderable set structures and in general from the axiom of choice for all set-sized structures. But to get the universal property that you mention for class-sized structures, mere AC is not enough, for one needs the global axiom of choice, which is the assertion that there is a proper class well-ordering of the universe. This does not follow from AC, for there are models of Gödel-Bernays set theory GB that have AC, but not global choice. But global AC is sufficient to carry out the embedding of any class linear order into No. A similar phenomenon arises with many other class-sized class-homogeneous set-saturated models, which require global AC to get the universality for class structures.

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Let $\mathcal{G}$ denote the class of all groups. Define

$$\mathcal{M} = \left\{ f : \mathcal{G} \dashrightarrow \bigcup\mathcal{G} : \left(\forall G \in \mathrm{dom}(f)\right)\left(f(G) \in G\setminus\{\mathrm{id}_G\}\right)\right\}$$

In other words, $\mathcal{M}$ consists of all set-sized partial choice functions on the class of all groups, excluding choice functions that ever pick the identity. The empty function is the identity of this class group, the inverse of a function in $\mathcal{M}$ is its point-wise inverse, and multiplication is defined as follows:

• $\mathrm{dom}(fg) = \mathrm{dom}(f) \cup \mathrm{dom}(g) \setminus \{G \in \mathcal{G} : f(G)g(G) = \mathrm{id}_G\}$
• $(fg)(G) = f(G)g(G)$ if $G\in\mathrm{dom}(f)\cap\mathrm{dom}(g)$ and $f(G)g(G)\neq\mathrm{id}_G$
• $(fg)(G) = f(G)$ if $G \in \mathrm{dom}(f)\setminus\mathrm{dom}(g)$
• $(fg)(G) = g(G)$ if $G \in \mathrm{dom}(g)\setminus\mathrm{dom}(f)$

The intuitive idea here is you want to take the Cartesian product of all the groups. In other words, your mother group would consist of all the class-sized choice functions on $\mathcal{G}$. Of course, a class can't have class-sized elements, so we try to settle for taking the class of all set-sized partial choice functions on $\mathcal{G}$. The problem with that boils down to what to take as the identity element? Or in other words, how do you differentiate a partial choice function from one that is similar, but additionally picks out some identity elements from groups in its domain? The solution is to "disallow" choosing the identity. This leads us to make the identity of this mother group the empty choice function, and to define multiplication essentially like point-wise multiplication, except that we remove any identities in the result.

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1. I presume you mean $f(G)$ above, not $f(g)$ (what's $g$)? 2. So was my supposition in the question post that the axiom of choice is required correct? –  mike3 Apr 1 '13 at 8:46
Yes, edited.${}$ –  Amit Kumar Gupta Apr 1 '13 at 9:32
Also, your construction of the direct limit of the symmetric groups makes sense in the context of choice, and also gives you a class group which contains an isomorphic copy of every set group. –  Amit Kumar Gupta Apr 1 '13 at 9:37