I have a homework question that asks:
Let a group G be generated by $\{a_i : i \in I\}$ where $I$ is an indexing set and $\forall i \in I,$ $a_i \in G$. Let $\phi : G\rightarrow G'$ and $\mu : G \rightarrow G'$ be two homomorphisms from $G$ into $G'$ such that $\forall i \in I $, $\phi (a_i) = \mu (a_i)$. Prove that $\phi = \mu$.
I am confused by the first sentence, does it mean that each $a_i$ is a generator of some subgroup of $G$ and we are looking at the union of those subgroups?
This is from section 13 of Fraleigh "A First Course in Abstract Algebra" problem 46. Section 13 is all about homomorphisms.
It means that all the elements of the group are generated by the $\;a_i\;$ , meaning:
$$\forall\,g\in G\;,\;\;g=b_1b_2\cdot\ldots\cdot b_m\;,\;\;\text{where}\;\;b_m=a_i^{\pm1}\;\;\text{for all}\;\;j=1,2,...,m$$
and $\;m\;$ is dependent on $\;g\;$ .