Is there a finite group whose identities are the same as the group identities?

Question

Consider the equational theory of groups in the signature $\{+,0,-\}$. The equational theory is axiomatized by the associative law, the identity laws, and the inverse laws. My question is, does there exist a finite group $G$ whose identities are precisely the same as the identities generated by the standard group identities? Note, I want $G$ to be finite.

Answer 1

Any finite group $G$, written additively, must satisfy $nx=0$ for $n=|G|$. The group $(Z;+,−,0)$ satisfies no such law. Hence the group $(\mathbb Z; +, -, 0)$ does not belong to any variety of groups that is generated by a finite group.

Also, it is known from the work of Birkhoff that if $V = {\sf H}{\sf S}{\sf P}(A)$ is a variety generated by a finite algebra $A$ of size $k$, then any $n$-generated algebra in $V$ will have size at most $k^{k^n}$. Hence, the finitely generated members of any finitely generated variety will be finite. This yields a sufficient criterion to show that a variety is not generated by a finite algebra: if the variety contains a finitely generated infinite algebra, then the variety cannot be generated by a finite algebra. For this problem, you can apply this criterion to $(\mathbb Z; +, -, 0)$, since it is finitely generated and infinite.