Image of Homomorphisms for Cyclic Groups I'm reading an example for abstract algebra and I would like to clarify some terms. The example is showing the number of homomorphisms from $\mathbb Z $ to $S_3$.
Take a homomorphism $f$ from $\mathbb Z$ to $S_3$. Since $1$ is a generator of $\mathbb Z$, $f$ is completely determined by the image of $1$. Since $1$ has infinite order, there are no restrictions on the image of $1$. Thus each element of $S_3$ is a possible image for $1$. There are $6$ homomorphisms from $\mathbb Z$ to $S_3$.
What does "completely determined by the image of $1$" mean? I thought about it a little bit, and it seems that since $1$ is a generator of $\mathbb Z$, any element in $\mathbb Z$ can be generated from $1^k$. When it says image of $1$, does it mean $f(1^k)$, where $k$ is an integer? In general, what does the "image" of an element under homomorphism mean? I thought for an element $g$ in group $G$, and for a homomorphism $h$, the image of $g$ under $h$ is $h(g)$. How can that determine the entire homomorphism?
Thanks for your help.

You are right. The image of $g\in A$ under a homomorphism $h\colon A\to B$ is the element $h(g)\in B$ to which $g$ is mapped. 
That $f\colon \mathbb Z\to S_3$ is completely determined by the image of $1$ means that if you have two homomorphisms $f_,g\colon \mathbb Z\to S_3$ and the image of $1$ is the same for both (i.e. $f(1)=g(1)$) then the homomorphisms are the same (i.e. $f(x)=g(x)$ for all $x\in\mathbb Z$).
More generally, if $A$ is a cyclic group with generator $a$ and $B$ is any group, then a homomorphism $f\colon A\to B$ is uniquely determined by the image of $a$. This follows from the homomorphism properties of $f$ and $g$ because $f(a)=g(a)$ gives us $f(a^{-1})=g(a^{-1})$ and then by induction $f(a^{\pm n})=g(a^{\pm n})$. Since $a$ generates $A$, this covers all elements of $A$. Even more generally, the corresponding statement holds also for groups with more generators.
