My question is the same as the one asked here
Group action defined on generators., namely given a group G=⟨ a,b∣R ⟩ where R is some set of relations. Then, the question defines a group action on X for each generating element.
The answers state that "What you need to check is that every relator will act (by concatenating the action for the generators as given in the relator) as the identity."
The answer mentioned then goes on to give an example of an action which satisfies the above sentence. However I am unable to rigorously prove that if the above holds then it is a group action. It would helpful if somebody could explicitly write down the proof. What troubles me is again how is 'ab' going to act on the set given the action of 'a' and 'b'. Proving it for any relation, example a^n will be a bit more helpful for me to understand.
Ok, let $I$ be any set and $R$ relations s.t $G=\langle I\mid R\rangle$, and assume that you have a map $f:I \to \mathrm{X}$ that respects $R$. Then the above claim follows immediately from the universal property of the free group and the definition of $\langle I\mid R\rangle$ as a quotient by the subgroup generated by $R$, so you can use these two univ properties to get an actionmap!
I.e. since you have a map $$\{I\} \to \mathrm{End}(X)$$ you get by the universal property of the free group an actionhomomorphism $$\varphi:\langle I\rangle \to \mathrm{End}(X)$$ which you can factor by the universal property of the quotient over $\langle I\mid R\rangle$ as $f$ respects $R$ we have $R \subset\mathrm{ker}(\varphi)$ and voila, you got an action group homomorphism $$\psi: G=\langle I\mid R\rangle \to \mathrm{End}(X)$$ which is by definition nothing else than an action of $G=\langle I\mid R\rangle$ on $X$