Let $G$ be a group and $S\subset G$ a generating set. Let $P$ (short for $P(x_1,\dots,x_n) = 1$) be an equational property that may or may not be satisfied by all $n$-tuples of elements of $G$. My question is:
What $P$ are such that $P$ being satisfied on $S$ implies that it is satisfied on all of $G$?
Is there a simple characterization? (If so, do you know a proof?) Is this a hard question with a developed theory? (If so, can you point me to relevant sources?)
Examples of $P$'s:
(1) If $P$ is $[x_1,x_2]= x_1x_2x_1^{-1}x_2^{-1}=1$, then $P$ being satisfied on $S$ implies that it is satisfied on $G$. Indeed if pairs of generators commute, then words on those generators clearly commute since the generators can "slip past each other."
(2) On the other hand, if $P$ is $x_1^2=1$, then $P$ being satisfied on $S$ says nothing about $G$. For an extremely hands-on example, $S_3$ is generated by two elements of order $2$, but not all its elements are order $2$.
Motivating example:
What got me interested in this in the first place was that I was wondering if free metabelian groups (on a finite number of generators) are finitely presented. Does the relation $[[x_1,x_2],[x_3,x_4]] = 1$ that defines metabelianness carry over from a generating set to the whole group? A few google searches seems to have revealed that it does not, so now I'm wondering what the other relations might be that do carry over.
Thanks in advance.
This is more of an extended comment:
If your group has two generators, like the free group on two generators, then these metabelian relations hold for the generators, since $[[x,y],[x,y]]=1,[[x,y],[y,x]]=1,[[x,y],[x,x]]=1$ (this is just asking if $[x,y]$ commutes with itself, with its inverse, and with the identity).
You should look into verbal subgroups and reduced free groups, I know that Combinatorial Group Theory by Karrass, Magnus, Solitar has some stuff on them. It is known that reduced free groups all have presentations $\langle a_1,...,a_n \mid X_1^d, P_1,....,P_i,...\rangle$ where the $X_1,P_i$ are all the words that satisfy the equation they specify (so $P=X_1X_2X_1^{-1}X_2^{-1}$ is the commutator subgroup and $X_1^d$ is the group generated by all $d$-powers), and $P_i$ are in the commutator subgroup (this is Cor 2.3.1 in Combinatorial Group Theory). You can find by looking directly at any original presentation and transforming the presentation in a fairly simple manner (although I am not sure if "defining relation on generators is enough"-property is preserved, but it seems plausible).
I wouldn't be surprised if the commutator equation is the only one where you can get away with generator only relations, and feel like there should be a way to finagle an answer using that corollary.
Update 1: The equation for the equation $[X,Y]X^p=1$ it is enough to define on generators (where $X,Y$ vary over the group.) This is because if you have generators $x_1,...,x_n$, you can look at $[x_i,x_i]x_i^p=1=x_i^p$, so your generators are torsion, and $[x_i,x_j]x_i^p=1=[x_i,x_j]$ (since $x_i^p=1$) so your group is commutative. So your group is $\langle x_1,...,x_n \mid x_i^p, [x_j,x_k]; 1\leq i,j,k \leq n \rangle $ which is exactly the free reduced group version (ranging over all elements rather than just the generators).
Note that if you change this to something like $[X,Y]W(X,Y)=1$ ,where $W(X,Y)$ is a word in $X,Y$ you still get that $x_i$ are torsion provided $W(X,X)$ is not in the commutator subgroup, by the same argument, but you don't know, a priori, that $W(x_i,x_j)$ will be trivial, so you won't be able to conclude $[x_i,x_j]=1$, which is important to the argument. The reduced free group version of this group will be torsion( provided $W(X,X)$ is not in the commutator group) using the techniques in the proof of corollay 2.3.1. It feels very unlikely if you got a group which ranged over just the generators you would end up with a torsion group for most choices of $W(X,Y)$.