Can a countably infinite torsion group $G$ exist in which $G=F\times\{0,1,2\}$ and $G^2=F\times\{1,2\}$?
OR (and perhaps this is a better description, but I don't quite know enough about countably infinite torsion groups just yet to discriminate whether this enhancement of the conditions is necessary):
Can a countably infinite torsion group $G$ exist in which $G=F\times\{0,1,2\}$ and $G^2=\Phi(F)\times\{1,2\}$?
Where $\Phi$ is a group isomorphism on $F$.
I can't really describe what I know and where I am stuck other than that my knowledge of group theory is very limited, but I will try:
Part of me says this is likely impossible because there is only one group of order $3$, and in that, every element is a square. But another part of me says I don't know enough to say that for definite, and there may be some way of combining $\{0,1,2\}$ other than a direct product, which works.
The answers here may be related.
I can however explain the motivation. This is a precondition for the conjecture that the Collatz graph on the odd integers is a map towards unity in a torsion group $G=\{2\Bbb Z+1,\circ\}$ in which $\circ$ has the property:
$x\circ x=(3x+1)\lvert 3x+1\rvert_2$
Where $\lvert\cdot\rvert_2$ is the 2-adic absolute value.
An answer in the affirmative would mean that if such a group can be constructed, it could be countable. There is an uncountably large extension of the Collatz graph to $\Bbb Z_2$ and I'm trying to determine a) whether this group-theoretic approach can be ruled out and b) whether a proof of this form can be attempted in $2\Bbb Z-1$ or whether the extension to $\Bbb Z_2$ will be required.
