Function related problem

Question

Let $g:N \to N$ be defined as

$g\left( {3n + 1} \right) = 3n + 2$

$g\left( {3n + 2} \right) = 3n + 3$

$g\left( {3n + 3} \right) = 3n + 1$, for all $n \ge 0$.

Then which of the following statement is true?

1. There exist an onto function $f:N \to N$ such that $f\circ g=f$.

2. There exist a one-one function $f:N \to N$ such that $f\circ g=f$.

3. $g\circ g\circ g=g$.

4. There exist a one-one function $f:N \to N$ such that $g\circ f=f$.

I tried to make $f(x)=x+1$, the first two condition will satisfy it but the third will will not do so as it will satisfy the condition $f(x)=x-2$.

Answer 1

I won't answer this question for you, but I'll try to get you started, because from what I can tell, you don't see what it's about.

The first couple of options ask about a function such that $f\circ g=f$. What would such a function be like? We'd have $$ \begin{align} f(3n+1)&=f(g(3n+1))=f(3n+2)\\ f(3n+2)&=f(g(3n+2))=f(3n+3)\\ f(3n+3)&=f(g(3n+3))=f(3n+1) \end{align} $$ Is there a one-to-one function or an onto function that satisfies these requirements?

For the other two parts, proceed as above. Start by unraveling the formulas to see what the question is really asking.