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The product $\times$ of two groups is associative and commutative and there's a neutral element $\{1\}$. Let's say I create "virtual groups" which are inverses with respect to $\times$ (like getting $\mathbb{Z}$ from $\mathbb{N}$). Then I have a group $G$ whose elements are all groups.

This isn't allowed, though, is it? Since a set can't be a member of itself.

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How is the product of 2 elements in $G$ defined? Maybe you can give an example of how you think this could work...and +1 for thinking out of the box. –  draks ... Jul 19 '12 at 6:20
Gromov famously constructed a «metric space of all groups» for a sensible meaning of «all». –  Mariano Suárez-Alvarez Jul 19 '12 at 7:32

3 Answers 3

up vote 11 down vote accepted

The Cartesian product is not commutative: if $G\ne H$, $G\times H\ne H\times G$, though these groups are isomorphic. It isn’t associative, either, because $\big\langle\langle x,y\rangle,z\big\rangle\ne\big\langle x,\langle y,z\rangle\big\rangle$, though once again there’s an isomorphism between $(G\times H)\times K$ and $G\times(H\times K)$.

That aside, there’s no problem in principle with having a group $G$ whose elements are themselves groups; $G$ simply won’t be one of those elements. What you can’t do is form a group $G$ whose elements are all possible groups, any more than you can form a set of all sets.

What you can do in ZF is write down a first-order formula $\varphi(x)$ that ‘says’ that $x$ is a group, meaning an ordered pair $\langle y,z\rangle$ such that $y$ is a set and $z$ is a function from $z\times z$ to $z$ with the properties needed to make it a group operation on $y$. You can then write down formulas $\psi(x,y,z)$ that ‘say’ that $x,y$, and $z$ are groups and that $z$ is related to $x$ and $y$ in some specific fashion. I shouldn’t be at all surprised if it were possible to find such a formula for which one could prove

$$\forall x,y\Big(\big(\varphi(x)\land\varphi(y)\big)\to\exists!z\psi(x,y,z)\Big)\;,$$

i.e., $\psi$ defines what would be a binary operation on groups if the set of all groups existed,

$$\forall u,v,w,x,y,z\left(\Big(\psi(x,y,u)\land\psi(y,z,v)\Big)\to\Big(\psi(u,z,w)\leftrightarrow\psi(x,v,w)\Big)\right)\;,$$

i.e., that ‘operation’ is associative, and the rest of the group axioms. Then one could talk informally about the proper class $\bf{G}$ of groups and a proper class operation making $\bf G$ into a group.

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However if one uses the equivalence classes of groups under isomorphism as members of this monoid of groups, you should get both associativity and commutativity. I'm not sure there can be a meaningful definition of inverse groups in this setting, though. –  celtschk Jul 19 '12 at 8:54
There is another problem here: even under appropriate equivalence classes, the product is not cancellative: there are nontrivial groups $A$ such that $A\times A\cong A$; and groups $A$, $B$, and $C$, pairwise nonisomorphic, such that $A\times B\cong A\times C$. In order to get a group by adding "formal inverses", we need to have a cancellation semigroup to begin with, otherwise we need a coarser equivalence relation than "isomorphism". –  Arturo Magidin Jul 19 '12 at 16:53
@ArturoMagidin, could you please give me an example of such a group? –  Koenraad van Duin Aug 9 '14 at 13:51
@ArturoMagidin, could you please give me an example of such a group $G$ that satisfies $G \times G \cong G$ –  Koenraad van Duin Aug 9 '14 at 13:59
@Koenraad: I was gone for almost a year, so I didn’t see this, and Arturo’s been gone for quite a while. An example is to pick any group $G$, let $G_n=G$ for $n\in\Bbb N$, and let $H=\prod_{n\in\Bbb N}G_n$; then $H\times H\cong H$. –  Brian M. Scott Jan 21 at 2:38

As Brian notes, there's no problem with a group whose elements are groups. Take your two favorite groups, let's call them $A$ and $B$, let $G=\{{A,B\}}$, and define a binary operation on $G$ by $$AA=BB=A,\quad AB=BA=B$$ Voila! a group whose elements are groups.

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In addition to the problems mentioned by Brian Scott, there is another problem.

You seem to be trying to mimic the construction of the enveloping group of a semigroup (this is the image of the adjoint functor to the forgetful functor from $\mathsf{Group}$ to $\mathsf{Semigroup}$ or to $\mathsf{Monoid}$); in the commutative case, this is often called the Grothendieck group. Generally, you do not have an embedding of the original semigroup into the enveloping; you need the semigroup to be cancellative for that to hold.

But in this setting, there are groups $A$, $B$, and $C$, pairwise nonisomorphic, such that $A\times B\cong A\times C$. That would mean that in the enveloping group, we would need to identify $A$ with the trivial group, or $B$ with $C$. You are going to have a lot of collapse in this resulting object. For example, if your class of groups includes the infinite direct product of copies of $\mathbb{Z}$, and this group is not identified with the trivial group, then every finite rank free abelian group will be identified with each other (a less than desirable outcome).

You could get away with doing this for, say, finite groups. You can even do it within ZFC, by restricting the groups to having underlying set contained in a particular infinite set. Then, under the equivalence relation of "isomorphism" you have a cancellative commutative monoid under cartesian product, so you can construct the Grothendieck group by adjoining "formal inverses". I don't think you get anything particularly interesting, though: the monoid is free abelian on the indecomposable groups, and so the Grothendieck group is just free abelian on the indecomposable groups.

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