Equivalence between Ising model and lattice gas model The Ising Hamiltonian is 
$$
H=-J \sum_{i=1}^N \sigma_i \sigma_{i+1} - H \sum_{i=1}^N \sigma_i
$$
How I can show the lattice gas model Hamiltonian is equivalent to that of the Ising model?
 A: The Lattice gas hamiltonian is:
$$H = A\sum_{i=1}^N  n_i + \frac{1}{2}\sum_{i,j} B_{ij}n_in_j $$
If we want to map this into Ising we just need to take a lattice such that the occupation number can only be $n_i=0,1$ (which is always possible) and then substitute:
$$n_i = \frac{1}{2} (1+S_i)$$
$S_i$ can assume values $-1$ or $+1$. The first term of the hamiltonian simply becomes:
$$\frac{A}{2}N + \frac{A}{2}\sum_{i=1}^N S_i$$
the second one instead is:
$$\frac{1}{8}\sum_{i,j} B_{ij} + \frac{1}{8}\sum_{i,j} B_{ij} (S_i+S_j) + \frac{1}{8}\sum_{i,j} B_{ij}S_iS_j $$
Assuming $B_{ij}=B_{ji}$ you can write $\sum_{i,j} B_{ij} (S_i+S_j) = 2\sum_{i,j} B_{ij}S_i$ by simply swapping the indices' names in the sum. Finally, assuming $B_{ij} = B$ for nearest neighbours and $0$ otherwise, the interacting term becomes:
$$\frac{BzN}{8} + \frac{Bz}{4}\sum_{i} S_i + \frac{B}{8}\sum_{\left\langle i,j \right\rangle} S_iS_j $$
(Here $z$ is the number of nearest neighbours). Putting everything together you get:
$$ H = \left( \frac{AN}{2}+\frac{BzN}{8} \right) + \left( \frac{A}{2}+\frac{Bz}{4} \right) \sum_{i} S_i + \frac{B}{8}\sum_{\left\langle i,j \right\rangle} S_iS_j $$
Now you set $h=-\left( \frac{A}{2}+\frac{Bz}{4} \right)$; $J=-B/8$ and you have Ising hamiltonian rescaled by a constant, which is not important for any thermodynamical property. 
