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The context of this question is from the definition of the sporadic Mathieu group $M_{23}$, which (in one possible definition) is the stabilizer of a point in $M_{24}$, which is a certain subgroup of $S_{24}$ (a permutation group on $24$ points). When I read this I was a bit surprised since they haven't specified which point is to be stabilized, but I assume that the content here is that it doesn't matter, since different stabilizers yield isomorphic subgroups.

Which brings me to the question: is it possible to define a single group given a set of isomorphic groups? Of course in this case we could define $M_{23}$ as the stabilizer of the first point, assuming that the $24$ original points are distinguishable to begin with, for example if they were the numbers $1,\dots,24$ and I could pick out $1$ specifically. But in more general contexts this can be problematic, especially for sets which have no distinguished members, such as the set of all free ultrafilters of $\Bbb N$.

If I have a family $(G_i)_{i\in I}$ of groups for some nonempty index set $I$ such that $G_i\cong G_j$ for all $i,j\in I$, is it possible to define a group $G$ such that $G\cong G_i$ for all $i$? By the observation above, if there is some term $x$ for which $x\in I$ is provable, then it would suffice to take $G=G_x$, but what if there is no such $x$?

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  • $\begingroup$ I'm a bit troubled by your saying "some term $x$ for which $x \in I$ is provable". If you're assuming $I$ is nonempty then $x \in I$ is provable, and you've answered your own question! And if there is no such $x$ then $I$ is empty, so there are no groups to consider and the problem is vacuous. $\endgroup$ – Clive Newstead Mar 16 '15 at 3:42
  • $\begingroup$ @CliveNewstead It's difficult to distinguish the notions in usual English, but there is a difference between $\vdash x\in I$ and $\vdash\exists x:x\in I$. Obviously the first implies the second, but the second does not imply the first, for example as I mentioned when $I$ is the set of free ultrafilters on $\Bbb N$ (there are free ultrafilters in $\sf ZFC$, but there are no definable free ultrafilters). $\endgroup$ – Mario Carneiro Mar 16 '15 at 5:00
  • $\begingroup$ @AndresCaicedo I hope I made it clear that the question has more to do with definability in a theory than groups themselves, and so I'm inclined to consider it more (set-theory) or some other foundational topic for its connections to models and provability than (elementary-set-theory), which seems to cover more explicit set-related topics like set operations and relations. $\endgroup$ – Mario Carneiro Mar 16 '15 at 5:04
  • $\begingroup$ @MarioCarneiro No, it is not set theory. $\endgroup$ – Andrés E. Caicedo Mar 16 '15 at 5:08
  • $\begingroup$ @AndresCaicedo Perhaps (logic) then. $\endgroup$ – Mario Carneiro Mar 16 '15 at 5:10
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Define in specific detail? Of course not. You can choose one from the collection.

But why choose? Why do anything at all? We can revisit the definition of cardinal in $\sf ZF$, as well cardinal arithmetic that goes along with it. Specifically, you might not be able to have a choice of representatives there.

Instead we abstract the notion of cardinal, we assign a certain set, using Scott's trick, to the equivalence class (proper class) of sets under equipollence. We then proceed to calculate addition and multiplication and so on, based on picking an arbitrary representatives from two classes, and showing that we can find a set with the wanted properties defined from those.

Here we can do the same thing. If $\{(G_i,\cdot_i)\mid i\in I\}$ is a set of groups which are pairwise isomorphic then we can say that they define a group $G$ which is abstract. In order to do some calculation in $G$ you pick an arbitrary $G_i$ to do the calculation in, and use the fact that there is an isomorphism between every two, so the result doesn't matter.

In some sense, this group is "pointless", and we only interested in knowing that it is isomorphic to those $G_i$'s, not what its points are. If you have two "pointless groups", $G$ and $H$ their product can be defined as the set $\{G_i\times H_j\mid i\in I,j\in J\}$ (where $\{ H_j\mid j\in J\}$ correspond to $H$, of course).

And just like in cardinal arithmetic over $\sf ZF$, if you already know that every group can be represented as a "pointless group", then you can just pick two representatives and consider the pointless group representing the product of these two groups. Of course, using Scott's trick you can define canonical pointless groups like that.

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  • $\begingroup$ I'd be happy to hear a verbal criticism as well! $\endgroup$ – Asaf Karagila Mar 16 '15 at 8:27
  • $\begingroup$ I see now what you mean by "an answer I may or may not be fond of". I wonder whether it is possible to prove that my construction is impossible, possibly by using the fact that it would have to be permutation-invariant under permutations of $I$ to extract explicit isomorphisms between the $G_i$. $\endgroup$ – Mario Carneiro Mar 16 '15 at 8:27
  • $\begingroup$ Mario, there are categories which are not concrete, and we cannot have this sort of choice of representative even if we wanted. This is not a big deal, here you just abstract the choice of representative to the meta-language and claim that it doesn't really matter for any useful purpose. It works for cardinal arithmetic without choice quite well. $\endgroup$ – Asaf Karagila Mar 16 '15 at 8:29
  • $\begingroup$ I know, but it tends to make for a lot of meta-mathematical hoops to jump through and makes formalization a pain in general, so I like to avoid it when possible, and I was hoping that $\sf Grp$ might admit such a counstruction. Your answer was well written, though (even if it has to take the disappointing negative stance). +1 $\endgroup$ – Mario Carneiro Mar 16 '15 at 8:32
  • $\begingroup$ Well, for this you just really need global choice, to choose one from each class. It's probably equivalent to global choice, too. $\endgroup$ – Asaf Karagila Mar 16 '15 at 8:33

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