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:

- they are ordered by inclusion,
- 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?