Question about AC

[Patrik Eklund]

Dear JS and MK, Dear all,

Let me share an experience with you.

We've been working with quantales, and now also finite quantales when we 
come to real-world applications. Doing that I realized we need and want 
to see all quantales for n=3,4,5,... . For n=3 there are 12 
(non-isomorphic) one, for n=4 129, n=5 1852, n=6 33391, and so on. We 
presently work to show all quantales for n=7 and n=8. For up to 6, the 
Catalogue is found at

   http://glioc.com/algebra/

and once we have worked out n=7 and n=8, I guess by June, that extension 
will be published at the same site. Going up to a bit larger n brings in 
new wishes, like now wanting to see dualizing elements based on 
residuation/implication.

Not going into particular applications, and why we do it, but the 
experience shared is that we can is some cases (smaller n) replace 
"seem" in "seem very rare" with something more precise (sometimes it's 
not as rare as we might think, or conversely). We also arrive at more 
practical understanding e.g. about products of quantales and modules 
involving quantales. As there are so many of them, and many look quite 
natural and useful, one starts to wonder what it looks like in real 
life, e.g. concerning sequences and sequencing. Are there naturally 
canonic ones, and if so, in what sense?

Doing so also makes one curious about things like "operators operate on 
operators" (can even be understood as term functors over Kleisli 
categories of other term monads; crazy idea, I guess, but why not?), and 
how such things appear in nature.

We've been using e.g. mace4 to generate the raw material, and then we 
add code to check for particular algebraic properties. This is indeed 
done on our side only for quantales, but can be done similarly with any 
structure of your choice.

What I say is that people seems (!) to work with finite structures 
without actually looking at them more broadly and comparatively. Which 
brings me to warmly recommend sometimes actually to look at them, all of 
them.

Best,

Patrik



On 2020-05-13 16:34, Joe Shipman wrote:
> Yes, but the motivating problem is to find a group  of order >=3 with
> no nontrivial automorphisms. If the factor groups are all order 3 and
> you take a direct product or direct sum, then sending every element to
> its inverse works. If the group is nonabelian it may not be a product
> but that isn’t needed because you can conjugate by an element not in
> the center.
> 
> I wonder how easy it is to classify groups whose automorphism group
> has order 2. For finitely generated groups this is very easy but there
> could be a lot of infinitely generated ones.
> 
> Automorphism groups of odd order seem very rare.
> 
> — JS
> 
> Sent from my iPhone
> 
>> On May 13, 2020, at 8:38 AM, katzmik at macs.biu.ac.il wrote:
>> 
>> I was thinking of a possible counterexample in the spirit of 
>> Russellian pairs
>> of socks (I hope the attribution is correct).  If one takes an 
>> infinite
>> collection of groups, each isomorphic to Z/3Z, but in such a way that 
>> the
>> non-dentity elements are not distinguished in any way, then the 
>> product of all
>> these groups will possibly give a counterexample.  Namely, a 
>> particular Z/3Z
>> surely has a complementary group but it seems to me in order to choose 
>> such a
>> complementary group one may need to invoke a weak form of AC.  But 
>> perhaps
>> there is a way of doing it differently.  MK
>> 
>>> On Wed, May 13, 2020 14:38, Joe Shipman wrote:
>>> Yes, sorry for the imprecision. I don’t need that H is a direct 
>>> summand in
>>> every case, just in the special case where every element has order 2.
>>> 
>>> The general isomorphism theorem that if H is a normal subgroup of G, 
>>> then
>>> every automorphism of H extends to an automorphism of G, also 
>>> requires choice,
>>> so I could have asked it that way too.
>>> 
>>> — JS
>>> 
>>> Sent from my iPhone
>>> 
>>>>> On May 13, 2020, at 2:27 AM, katzmik at macs.biu.ac.il wrote:
>>>> 
>>>> You should formulate your question more carefully because if G=Z_4 
>>>> and
>>>> H=Z_2
>>>> then H is a subgroup of G but there does not exist K such that G is
>>>> isomorphic
>>>> to H x K.
>>>> 
>>>> MK
>>>> 
>>>>> On Tue, May 12, 2020 20:25, Joe Shipman wrote:
>>>>> Let G be an abelian group and H be a finite subgroup of G.
>>>>> Is some form of AC necessary to prove that there exists a group K 
>>>>> such that
>>>>> G
>>>>> is isomorphic to H x K ?
>>>>> 
>>>>> Motivation: the theorem “If |G| >= 3, then |Aut(G)| >= 2” is proved 
>>>>> by
>>>>> cases—
>>>>> If G is not abelian, conjugate by an element not in the center of G
>>>>> If G is abelian and some element is not its own inverse, send every 
>>>>> element
>>>>> to
>>>>> its inverse
>>>>> If G is abelian and every element is its own inverse, and |G|>=3, 
>>>>> there is
>>>>> a
>>>>> subgroup H isomorphic to Z2xZ2, express G as HxK and extend one of 
>>>>> the
>>>>> automorphisms of H.
>>>>> 
>>>>> But this last case required K existing with G isomorphic to HxK, 
>>>>> does that
>>>>> need Choice?
>>>>> 
>>>>> — JS
>>>>> 
>>>>> Sent from my iPhone
>>>>> 
>>>> 
>>>> 
>>> 
>>> 
>> 
>>