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Let $A$ be a set, together with a set $F$ of n-ary operations on A, which may include constants of $A$ as 0-ary operations. A set $G$ of operations on $A$ is said to be auxillary with respect to the algebra $(A,F)$ if $G$ is disjoint from $F$ and the equational identities of $(A,F)$ generate the equational identities of $(A, F \cup G)$. For example, in $(\mathbb R, +)$, every {$r$} where $r$ is a nonzero real, is an auxillary set. My question is, take a group $G$ in the signature {$*$}. Suppose $g$ is not an n-th root of unity. Is {$g$} an auxillary set with respect to $(G, *)$. I apologize if the question is too long, but I would be VERY surprised to see a counterexample.

user107952
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  • Perhaps if you take $g$ in the center of the group, the identity $\forall x: gx = xg$ would not follow from the identities of $G$. But it seems that the same reasoning would work for a root of the unity. For instance, why does $g^{2n}=g^n$ follows from the identities of $G$? – Pedro Sánchez Terraf Mar 25 '16 at 14:13
  • I am not sure how to interpret your question, since in universal algebra, a group is usually defined with the signature $(*, 1, {}^{-1})$. – J.-E. Pin Mar 29 '16 at 11:12
  • @J.E.Pin I mean the reduct in the signature {*}. – user107952 Mar 31 '16 at 04:32
  • @PedroSánchezTerraf For example in the multiplicative group of the reals, (x * -1 * -1) = x does not follow from the identities of the multiplicative group. The only identities are those that can be generated by the associative and commutative identities. – user107952 Apr 12 '16 at 23:56
  • So, a non-root-of-the-unity element of the center of a non-commutative group should be a counterexample (as I indicated above). Please correct me if I'm wrong. – Pedro Sánchez Terraf Apr 13 '16 at 11:59

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