Question about AC

[katzmik at macs.biu.ac.il]

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
>