Question about AC

[katzmik at macs.biu.ac.il]

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
>>>
>>
>>
>
>