Well definedness of a map defined on a set of generators of a group G.

Question

Let $G$ be a group with a presentation $(S,R)$, where $S$ is the set of generators and $R$ is the set of relations. I want to define a map from $G$ to some group $H$. I know where elements of $S$ goes. So naturally I can extend to whole of $G$. How to check that the extended map is well defined?

Answer 1

It is well defined if and only if the images of the elements os $S$ satisfy the same relations $R$ as the group $G$ itself. So, for instance, if you have a single ralation $xy=y^2x$ and if your map is $f$, the equality $f(x)f(y)=f(y)^2f(x)$ must hold.

Answer 2

Let $G$ be the group $ \langle S | R \rangle $. Let $H$ be any group. Given any map $S \to H$ given by $s \mapsto \bar{s}$, there exists a morphism $G \to H$ extending it if and only if the relations of $R$ are satised by the elements $\bar{s}$, i.e. if for any word $r(s_1, . . . , s_m)$ in $R$, the product $r_H(\bar{s}_1, . . . , \bar{s}_m)$ is trivial in $H$.

the proof is: Denote by $\pi$ the morphism $F(S) \to G$ extending $S \to G$. Suppose there exists a morphism $\phi : G \to H$ extending $s \mapsto \bar{s}$. Then $r_H(\bar{s}_1, . . . , \bar{s}_m)=\phi (r_G(s_1, . . . , s_m))=\phi(1)=1$ since $\phi$ is a morphism.

For the other direction: let φ : F(S) → H be the unique morphism extending $s \mapsto \bar{s}$. For any $r(s_1, . . . , s_m) ∈ R$, we have that $φ(r(s_1, . . . , s_m)) = r_H(\bar{s}_1, . . . , \bar{s}_m) = 1$ by hypothesis. Thus Ker(φ) contains $R$, and it is normal - it must thus contain $\langle \langle R \rangle \rangle = Ker(π)$. Hence φ factors through π i.e. $φ = φ' \circ π$. The morphism $φ' : G → H $ extends $s \mapsto \bar{s}$.