Möbius function and Posets

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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.

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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?

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$(i)\Rightarrow (ii)\forall f,g\in I(P): (f.g)(a,a)=f(a,a)g(a,a)=g(a,a)f(a,a)$, 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 $[a,b]={a}$.