Möbius function and Posets As usual, $\mathbb{C}$ denote the field of complex numbers.  Let $\mu \in I_{\mathbb{C}}(P)$ (the $\mathbb{C}$ incidence-algebra of $P(X, \leq)$ a poset). I am asked to show the following are equivalent.
(i) Given $a, b \in X$, we have $a \leq b \iff a = b$.
(ii) $I_{\mathbb{C}}(P)$ is a commutative ring.
(iii) There is a positive integer $n$ such that $\mu^n = 1$.
I am pretty lost on this and any help would be greatly appreciated. 
 A: HINT: The approach should be showing $(i)\rightarrow (ii)\rightarrow (iii)\rightarrow (i).$
For the first implication, if $(i),$ then how many terms are in the convolution? See that your field is commutative.
For the second implication try to show contrapositivity, what happens if $\mu ^ n\neq 1.$ What can you say about $\zeta $? Does $\zeta$ commutes with an arbitrary $\alpha$?
For the third implication, If $\mu ^n=1$ what can you say about $\zeta ^n$?  What is $\mu ^n$? For example, notice that $\zeta ^2(a,b)$ is the number of elements in between $a$ and $b.$ So for the equality to mantain, how many elements should be in every interval? 
A: $(i)\Rightarrow (ii)\forall f,g\in I(P): (f.g)(a,a)=f(a,a)g(a,a)=g(a,a)f(a,a)$, as $g(a,a),f(a,a)$ are in $C$ which is a field. 
$(iii)\Rightarrow(i)$: since $\xi ^{n}(a,b)=no.\begin{Bmatrix}
(x_{1},....,x_{n-1}):a\leq x_{1}\leq ...\leq x_{n-1}\leq b
\end{Bmatrix}$, and $\mu^{n}=1\Rightarrow \mu^{n-1}=\zeta \Rightarrow \zeta ^{n}=\mu^{n(n-1)}=1$. Hence $\zeta ^{n}(a,b)=1\Rightarrow \left |\begin{Bmatrix}
(x_{1},....,x_{n-1}):a\leq x_{1}\leq ...\leq x_{n-1}\leq b
\end{Bmatrix}  \right |=1\Rightarrow
 x_{i}=a,\forall i.$ Hence $[a,b]={a}$.
