Question about AC

[Joe Shipman]

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

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