# Can a countably infinite torsion group $G$ exist in which $G=F\times\{0,1,2\}$ and $G^2=F\times\{1,2\}$?

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.

No. $G^2$ contains the identity while $\{1, 2\}\times F$ does not. (I am assuming that $\{1, 2\}\times F$ is a notation for $G\setminus F$, that is, for the set $\{(1, f), (2, f)\mid f\in F\}$. This is a union of two of the three cosets of $F$ in $G$.)
Indeed, $F^2\subseteq F$ and so $G^2\cap F$ is quite big! However, $F\cap (\{1, 2\}\times F)=F\cap (G\setminus F)$ is empty.
• Thank-you. I may have carelessly implied $0$ is the identity and $+$ is the operation on $\{0,1,2\}$ but I meant to leave that vague and permit selection of any two elements out of the three. Commented Jun 7, 2018 at 10:58
• If $0$ is not the identity, but instead $1$ or $2$ is then the answer is still no. I will leave you to figure out why. Commented Jun 7, 2018 at 11:24
• Thanks; that's what I thought. My thinking is that it's a straightforward consequence of the fact that there's only the one group of order $3$ and $(Z3)^2=Z3$. In fact I think that would give us the theorem that no group which is a direct product of $\{0,1,2\}$ can be torsion with exponent $2^p$ for any $p$. Are there other candidates for arriving at the $2/3$ rule I seek which I might I try using $\{0,1,2\}$, other than a direct product? Commented Jun 7, 2018 at 12:05