Question about AC

[Joe Shipman]

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
>> 
> 
>