# The Nim Group is a group?

Consider the following exercise.

Let the set $G = \{0, 1, 2, 3, 4, 5, 6, 7\}$. Suppose there is a group (Called The Nim Group) operation $\ast$ on $G$ that satisfies the following two conditions:

(a) $a\ast b \geq a + b$ for all $a, b \in G$,

(b) $a\ast a = 0$ for all $a \in G$.

Construct the multiplication table for $G$.

The exercise I found him in a document but I have a doubt if indeed the set $G$ with the operation defined as a group.

I have a doubt with this exercise. Building the table until I get the following

My question is not how to continue, as should repeat elements in rows (being $G$ a group) then it must meet that $1 \ast 7 = 2$ but if this is true not satisfy the condition (b) $a\ast b \geq a + b$. I appreciate your support.

• Are you sure (a) is supposed to be regular addition, which suffices to fill out the whole table? I believe it should be bitwise XOR, in which case (b) is satisfied but redundant. I don't see how (b) gives you $a*b \ge a+b$-we don't have any order on this group. – Ross Millikan Sep 18 '13 at 17:00
• With your new condition (a), how can $6 * 7 \ge 13$? – Ross Millikan Sep 18 '13 at 17:01
• The group operation in NIM is usually defined recursively as follows $$ab=\operatorname{mex}\left(\{a'*b\mid a'<a\}\cup \{ab'\mid b'<b\}\right).$$ Here $\operatorname{mex}(S)$ is defined for all proper subsets of $\mathbb{N}$ and means the smallest non-negative integer not in the set $S$ (=Minimum EXcluded number). So $0*0=0$ simply because both sets on the r.h.s. are empty. But then $0*1=\operatorname{mex}(\{0\})=1=1*0$, $1*1=\operatorname{mex}(\{1\})=0$, $0*2=\operatorname{mex}(\{0,1\})=2=2*0$, $1*2=3$, $2*2=0$ et cetera. – Jyrki Lahtonen Sep 18 '13 at 17:01
• Can you name/link the document you found this in? – Mark S. Nov 16 '13 at 16:38
• @RoinerSeguraCubero I think (a) should be $a*b\leq a+b$ – zed111 Feb 3 '15 at 17:49

So the question in the book is exactly the same as you wrote, but instead of your first condition, the condition in the exercise is $a*b\leq a+b$, which makes more sense as you pointed out.