# Why the axiomatic rank of the variety of groups is equal to three?

I am thankful of Anton Klyachko who introduced axiomatic rank to me: the axiomatic rank of a variety is the minimum number of variables which we need to define that variety by identities. It seems clear that the axiomatic rank of the variety of groups is equal to three. But why?

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If I read en.wikipedia.org/wiki/Alternative_algebra and en.wikipedia.org/wiki/Octonion correctly, unit octonions are a nonassociative Moufang loop where the subalgebra generated by any pair of elements is associative, and therefore, hopefully, a subgroup. – Emil Jeřábek Jan 20 '14 at 18:54

Let $V$ be the variety axiomatized by \begin{align*} x+y&=y+x,\\ x+(x+y)&=y,\\ x+0&=x,\\ -x&=x. \end{align*} (This also implies $x+x=0$.) Straightforward induction on the length of a term shows that every $2$-variable term is equal over $V$ to one of $0,x,y,x+y$, which implies that $V$ axiomatizes all $2$-variable identities valid in abelian groups of exponent $2$. If we can find a nonassociative algebra $A\in V$, it will follow that no variety of associative algebras (in the signature of groups) containing the group of order $2$ (which generates the variety of abelian groups of exponent $2$) can be axiomatized by identities in less than $3$ variables.
I can’t think of a clever way to construct $A$ at the moment, but the following (which is just the $3$-variable term algebra modulo the theory of $V$ in slight disguise) works.
Let $A$ consist of a special element $0$ together with all finite unordered binary trees whose leaves are labelled $x$, $y$, or $z$, such that the two children of any non-leaf node are distinct (nonisomorphic), and if $a$ has children $b,c$, and $c$ is not a leaf, then its children are distinct from $b$. Define an operation $+$ on $A$ by $t+0=0+t=t$; $t+t=0$; if $s$ is a child of $t$, then $t+s=s+t$ is the other child of $t$; in the remaining cases, $t+s$ is the tree with a root whose children are $t$ and $s$. Then one can check that $A\in V$, and it is clearly nonassociative.
Your octonion argument is much simpler and works perfectly. The algebra of all unit octonions or all non-zero octonions (with operations $\cdot$, ${\ }^{-1}$, and $1$) is non-associative but satisfies all two-variable laws that are consequences of the group axioms. – Anton Klyachko Jan 20 '14 at 21:26
The point is that the argument here applies not just to the variety of all groups, but to every variety of groups that contains the two-element group. In fact, I think it can be modified so that it works with any prime exponent $p$ instead of $2$, which would show that every nontrivial variety of groups has axiomatic rank at least $3$. The octonions are a much simpler structure, but they do not give this level of generality. – Emil Jeřábek Jan 20 '14 at 21:56