Identities satisfied by free structures 
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*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 $G$?


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


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


*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!)
 A: *

*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.

*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.

*(1) and (2) are not true for rings or abelian groups.

*(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.
