# Order of g in Paillier Scheme

I'm trying to understand the Paillier Scheme but there's something I can't understand in the keyGen algorithm:

Ensure $$n$$ divides the order of $$g$$ by checking the existence of the following modular multiplicative inverse: $$\mu = (L(g^\lambda \bmod n^2))^{-1} \bmod n.$$

I'm trying hard to find the relation between $$n$$ divides the order of $$g$$ and $$\gcd(n, L(g^\lambda \bmod n^2))=1$$ but I can't find it. I hope that somebody can help me understand it.

PROPOSITION 13.6 Let $$N=p q$$ , where $$p, q$$ are distinct odd primes of equal length. Then:
1. $$\operatorname{gcd}(N, \phi(N))=1.$$
2. For any integer $$a \geq 0,$$ we have $$(1+N)^{a}=(1+a N) \bmod N^{2}.$$
As a consequence, the order of $$(1+N)$$ in $$\mathbb{Z}_{N^{2}}^{*}$$ is $$N .$$ That is, $$(1+N)^{N}=1 \bmod N^{2}$$ and $$(1+N)^{a} \neq 1 \bmod N^{2}$$ for any $$1 \leq a
1. $$\mathbb{Z}_{N} \times \mathbb{Z}_{N}^{*}$$ is isomorphic to $$\mathbb{Z}_{N^{2}}^{*},$$ with isomorphism $$f : \mathbb{Z}_{N} \times \mathbb{Z}_{N}^{*} \rightarrow \mathbb{Z}_{N^{2}}^{*}$$ given by $$f(a, b)=\left[(1+N)^{a} \cdot b^{N} \bmod N^{2}\right]$$
It means that $$n$$ divides the order of $$g=f(a,b)$$ iff $$L(g^\lambda \bmod n^2)=a\cdot \phi(N)\in \mathbb{Z}_{N}^{*}.$$