Question about AC

[Joe Shipman]

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
>>>> 
>>> 
>>> 
>> 
>> 
> 
>