Question concerning how a map extends to a homomorphism. Let $G$ be a finite group with finite set of generators $g_1,g_2,...$ that is not known "explicitly" as a group, but rather indirectly as a faithful representation as a subgroup of $S_n$. If I define a map $f$ from $G$ to another group $H$, that is defined on the generators $g_i$ satisfies $g_i^t=1$ then $(f(g_i))^t=1$ and I do not have a full set of relations, but just know the group as a subgroup in $S_n$, is there something I can do to check whether $f$ can be extended to a well defined homomorphism, that does not require checking if $f(ab)=f(a)*f(b)$ holds for all elements $a,b$? So does there exist a smaller subset I can check this equation on to ensure it holds on the whole of $G$? Or can always no matter how big this proper subset is, something go wrong?
As an example, suppose I know the orders of the pairwise products $(g_ig_j)^k=1$ and I check $(f(g_i)*f(g_j))^k=1$ does this imply $f$ can be extended as a well defined homomorphism? (I assume not)
 A: A collection of relations $R_1(g_1,g_2,\ldots),R_2(g_1,g_2,\ldots),\ldots$ in the generators is sufficient for your check (that is: it suffices to verify that $R_1(f(g_1), f(g_2),\ldots)=1$ etc.) if and only if the group presentation $\langle \,x_1,\ldots,x_n\mid R_1(x_1,x_2,\ldots), R_2(x_1,x_2,\ldots), \ldots\,\rangle$ is isomorphic to $G$ (via $x_i\mapsto g_i$). 
Hence only verifying orders of the $g_i$ is not suffifiecnt (unless you have only one generator) because groups $\langle\,x_1,\ldots,x_k\mid x_1^{n_1}=x_2^{n_2}=\ldots=x_k^{n_k}=1\,\rangle$ are infinite (if $k\ge2$) and your $G$ is finite.
Only verifying orders of pairwise products will also not be sufficient except in trivial cases. It is hard to tell when a set of relations is sufficient, but one pleasent situation is when the relations allow you to rewrite an arbitrary word in the generators (and their inverses!) into a standard order (e.g. always into the form $g_1^{a_1}g_2^{a_2}\ldots g_n^{a_n}$), for example if you have relations $g_ig_j=g_j^ag_i^b$ for all $i>j$ - which is not possible for all $G$.
