$\require{begingroup}\begingroup \DeclareMathOperator{\Var}{Var}$ Let $N_G$ denote the number of copies of a graph $G$ in the Erdős–Rényi random graph model $G(n,p)$. We have the law of large numbers for the number of copies of of graph $G$, i.e., $N_G$ is very close to the expectation $EN_G$. This means that whenever the expectation tends to zero, $N_G$ also tends to zero and we have $$\frac{N_G - EN_G}{EN_G} \overset{p}{\rightarrow} 0.$$ Now we are interested to know whether the central limit theorem is also true: i.e., \begin{equation} \cfrac{N_G-EN_G}{\sqrt{\Var N_G}}\overset{d}{\rightarrow}\mathcal{N}(0,1), \end{equation} for $\mathcal{N}(0,1)$ the standard normal distribution.
As with the Poisson limit theorem for the number of copies of $G$, this fact can also be proved by the method of moments. So if we can prove that if we have a sequence of random variables and for any $k$, the $k$th moment of the sequence tends to the $k$th moment of a standard normal random variable, then this convergence in distribution also holds.
To prove this, we need to prove that the $k$th moment of $E(\frac{N_G - EN_G}{\sqrt{\Var N_G}})^k$ is equal to $\frac{E(N_G - EN_G)^k}{(\sqrt{\Var N_G})^k}$ when $k$ is even. To do this we use the indicator function and the collection of $k$ vertices that can be pairwise disjoint or maybe not; i.e., maybe the first collection and the second are not disjoint or the the second and third and $\dots$ and maybe all of them are not disjoint!
But we can use some notation in this case that maybe will help me. We can consider a graph of relations between our collections and we can say that two collections are related to each other when they are not disjoint. So we can consider a graph on $k$ vertices such that each vertex is one of our collections and two vertices are adjacent when their collection do intersect (have nonempty intersection).
And it turns out that there are three situations here for such a graph $H$ on $k$ vertices with the above-mentioned specification.
In fact we have a lot of graphs here, i.e., $2^{{k \choose 2}}$ graphs, but we have only three different situations:
$H$ has an isolated vertex.
$H$ is a perfect matching.
else.
Now let me put what I said in equations to make it possible to understand the proof of the second case better:
Let's drop the denominator. We have \begin{eqnarray} E(N_G-EN_G)^k &=& E(\sum_i(I_i-EI_i))^k\\ &=&\sum_{i_i,\ldots,i_k\in I}E((I_{i_1}-EI_{i_1})\cdots(I_{i_k}-EI_{i_k})) \end{eqnarray} where $N_G=\sum_iI_i$.
Now $i_j\sim i_k \iff i_j\cap i_k \ne \emptyset$ so we get \begin{equation} =\sum_{H:V(H)=k}\sum_{\overset{(i_1,\ldots,i_k)}{j\sim h \text{ in } H \iff i_j\sim i_h}}E(I_{i_1}-EI_{i_1})\cdots (I_{i_k}-EI_{i_k}) \end{equation}
We can prove the first case as follows:
Let $i_1$ be an isolated vertex. Then $I_{i_1}$ is independent of $(I_{i_2}, \cdots , I_{i_k})$. Then $$E((I_{i_1} - EI_{i_1}) \cdots (I_{i_k} - EI_{i_k})) = E(I_{i_1} - EI_{i_1}) E(I_{i_2} - EI_{i_2}) \cdots E(I_{i_k} - EI_{i_k}) = 0$$ and we are done. This means that for the first case we have $o((\Var N_G)^{\frac{k}{2}})$.
Now we need to prove that the sum in the second case converges to $(k-1)!!\Var(N_G)^{k/2}$.
But to prove the second one I need help.
Thanks! $\endgroup$