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Are there any combinatorial games with odd order (under the usual addition of combinatorial games), apart from $0$?

In Are there combinatorial games of finite order different from $1$ or $2$? I asked about games of finite order greater than $2$, and was given a really nice example of a game of order $4$ (and I feel that the given example can probably be generalized to get games whose orders are higher powers of $2$), but in the comments to that answer it was indicated that there might not be any of odd order.

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No, there are no (non-zero) games of odd order, but you can indeed construct games whose order are arbitrary powers of two. There is a beautiful proof of these facts in "Combinatorial Game Theory" by A. Siegel, chapter III section 3.

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    $\begingroup$ The proof of "no (nonzero) games of odd order" is too long to fairly reproduce here. I suppose it's worth mentioning that the key result (which is not so easy to prove) is that if $G$ has finite order and birthday $n$, then $2^nG=0$. $\endgroup$ – Mark S. May 26 '14 at 15:22

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