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  1. Let $F$ be a free group. Does it follow that $F$ has the following property: for any $\{e,\circ, ^{-1}\}$-identity $s\approx t$ (in the sense of universal algebra), $F\models s\approx t$ if and only if $G\models s\approx t$ for all groups $G$?

  2. If $F$ has the above property, does it follow that $F$ is free?

  3. Is the answer to 1. and 2. the same for all kinds of algebraic structures: abelian groups, rings, ...?

  4. If it is false for some, is it at least true for kinds of algebraic structures defined without axioms, such as magmas?

My motivation for these question is that I heard the vague statement that a free group is a group satisfying all equations forced by the group axioms but no other, and 1. is a way of making this idea precise. (But I'm not sured if that is meant!)

user959368
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    The following are relevant: (i) https://math.stackexchange.com/questions/3824541/what-is-the-name-of-this-kind-of-algebra/3824723#3824723, (ii) https://math.stackexchange.com/questions/4006224/a-free-algebra-that-is-not-generic – Keith Kearnes Aug 17 '21 at 19:02

1 Answers1

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  1. No. Consider the zero group, which is the free group on 0 elements. The zero group satisfies $\forall x (x = e)$, but this identity is not satisfied by all groups.
  2. No. Consider the group generated by the infinite family of variables $\{x_i \mid i \in \mathbb{N}\}$ and also by the variable $y$, with the identity $y = y \circ y$ enforced. This group has the free group on $\{x_i \mid i \in \mathbb{N}\}$ as a subgroup (in fact, as a retract). Thus, any identities satisfied by this group are also satisfied by the free group on $\mathbb{N}$, and hence by all groups. But $y$ has order 2, and in a free group, all non-identity elements have infinite order.
  3. (1) and (2) are not true for rings or abelian groups.
  4. (1) and (2) are also not true for magmas.

Here is what is meant:

Consider the free group/ring/... on $n$ elements with generators $y_1, ..., y_n$. Consider any potential identity $t \approx q$ where $t, q$ are terms build from variables $x_1, ..., x_n$ and the group/ring/... operations. Then this identity holds when substituting $y_i$ for $x_i$ in the free group/ring/... if and only if it holds for all groups/rings/...

This follows from the definition of a free algebraic structure.

Mark Saving
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  • Thanks! Is there also a nontrivial counterexample to 1 (with more than one generator), and similarly for 3. and 4.?` – user959368 Aug 18 '21 at 11:25
  • @user959368 There is no nontrivial counterexample to 1 because two non-abelian free groups satisfy the same first-order theory. For any algebraic structure, you can take the free structure over 0 elements, which always satisfies $\forall x \forall y (x = y)$ (but this is generally not satisfied for all algebraic structures). Coming up with counterexamples for 2 is less trivial, but not too difficult. You basically follow the same idea as for free groups - embed the free structure on $\mathbb{N}$ elements into a generated structure on $\mathbb{N} + 1$ elements with a nontrivial identity. – Mark Saving Aug 18 '21 at 15:28
  • Thanks! So you are claiming that all non-trivial free structures (with more than one element) satisfy only those identities which are satsified by all structures (of the same type)? – user959368 Aug 19 '21 at 11:44
  • @user959368 To be precise, consider the $n$ generators of the free structure on $n$ elements. An identity on these $n$ elements is satisfied if and only if the identity is satisfied for all structures and any $n$ elements therein. But note that if we have more than $n$ variables in the identity, this need not happen. This is why I started with the zero group (free group with 0 generators) and gave an identity involving one variable. – Mark Saving Aug 19 '21 at 15:20