"If $\langle g\rangle=G$ then the image of the homomorphism $f:G\to H$ is determined by $f(g)$, $\operatorname{Im}(f)=\langle f(g)\rangle$,but determining the image of the generator does not define a homomorphism."
The first statement says that if a group is cyclic then the homomorphism is defined by the image of the generator but not the opposite?
Best Answer
If $G=\langle g\rangle$ then, if $f_1,f_2\colon G\longrightarrow H$ are group homomorphisms such that $f_1(g)=f_2(g)$, then $f_1=f_2$. That is, a homomorphism from $G$ into $H$ is totally determined by the element of $H$ that $g$ is mapped into.
However, if $h\in H$, it is not necessarily true that there is a group homomorphism $f\colon G\longrightarrow H$ such that $f(g)=h$. That is what happens if, for instance, $G=\mathbb Z_2$, $g=[1]$, $H=(\mathbb Z,+)$, and $h\neq0$.