Monoid morphisms For one of my computer science classes, I am asked to give examples of monoid morphisms for the following morphisms.
 
However, I really don't know how to approach it since we haven't worked with monoid morphisms.
If I understand correctly, in exercise 1, I should be giving an example of a mapping between $\{a,b,c\}^*$ and $(\Bbb N,+,0)$. 
For the 2nd exercise, I should be mapping $\big(\Bbb N \setminus\{0\},\cdot,1\big)$ and $(\Bbb N,+,0)$. 
Thanks in advance.
 A: For your first question: 
The monoid $\left\{a,b,c\right\}^*$ is usually referred to as the monoid "freely generated" by $a$, $b$, and $c$. What that means is that to define a monoid morphism out of it, you really just have to chose an image at your will for $a$, $b$ and $c$, and this will completely determine what you chose for the rest. Let me give an example when you only have the monoid $\left\{a\right\}^*$. Suppose that I decide to set $a\mapsto n$ for a given $n\in\mathbb{N}$, then since you want the map to be a monoid, you necessarily have that $aa\mapsto n+n$, and $aaa\mapsto n+n+n$. In general, you can show that for the word $w_{l_a}$ composed of $l_a$ times the letter $a$, you have $w\mapsto n\times l_a$. Conversely, these are indeed monoid morphisms. If you have understood this example, you can try as an exercise to generalize it to the case with $3$ generators, chosing an image for $a$, $b$ and $c$ and trying to figure out the image of any word in $\left\{a,b,c\right\}^*$
For your second question : 
I think you need to understand the first question first, since it is definitely easier, but then you can do a pretty similar reasoning by noticing that any non-zero number can be written as a product of primes in a unique way, up to the order. So the monoid $(\mathbb{N}\backslash0,\times,1)$ morally looks like a monoid freely generated by the prime numbers, and you can try chosing an image for any prime number. (To be precise it is the free commutative monoid on the set of prime numbers, but fortunately the monoid you land in is also commutative, so everything works out in the end).
