Is there a combinatorial game which is incomparable with 0 and every order 2 game? I've recently been playing around with combinatorial games and I came across what I've been calling halves of games. In particular, if $G$ is a game then a game $x$ is a half of $G$ if $x+x=G$.
Clearly if $x$ is a half of $G$, then so is $x+y$, $\forall y \in R$, where $R$ is the group generated by all order 2 games. In fact, this produces all halves of $G$. So there isn't immediately a good notion of the half of G.
It occurred to me that perhaps a way forward is comparing $G$ and its halves with the members of $R$. If there were a unique half of $G$, call it $x$, such that $x$ is comparable with the same members of $R$ as $G$ is, then it might be a good notion of the half of $G$. It should be mentioned that the members of $R$ are mutually incomparable, so comparability with one member of $R$ is in a sense independent of comparability with any other member of $R$.
That got me thinking about which games $x$ were comparable with which members of $R$. Clearly if $x \in R$, then the only member of $R$ which is comparable with $x$ is $x$ itself. Every game I can think of is comparable with at least one member of $R$, but I can't seem to prove that all games are comparable with at least one member of $R$, despite $R$ being vast. In fact, $\{x|-x\} \in R$, is distinct for all ordinals $x$.
This finally brings me to my question. Is there a game $G$ which is incomparable with every member of $R$? If there is, is there also a game $G$ with finite birthday which is incomparable with every member of $R$? If not, is there a general construction which takes a game $G$ and builds a member of $R$ which is comparable with $G$?
 A: I wasn't expecting to answer my own question when I wrote it, but I subsequently have. There are in fact many combinatorial games which are incomparable with $0$ and all games of order $2$.
It turns out that the whole of the torsion subgroup $T$ of the group of games has the property that distinct games in the subgroup are incomparable. Suppose to the contrary that $x,y \in T$ and $x \neq y$, but $x$ is comparable with $y$. Then without loss of generality, take $x < y$. Then $x,y$ have positive integer orders $m,n$ respectively. Then summing $mn$ copies of $y-x$ gives $0$. But from $x < y$, we get $y-x > 0$, therefore $mn(y-x) > 0$ and hence $0 > 0$, a contradiction.
Moreover, there are games of order $4$ with finite birthdays and therefore games with finite birthday which are incomparable with 0 and all order 2 games. For example, $\{1+*|-1\}$ has birthday $3$ and order $4$. It is incomparable with all games of order $1$ or $2$.
This answer does of course immediately pose another question, which was perhaps a better question to ask. Are there any games which are incomparable with all torsion games? I don't currently have an answer.
