How to arrive at the Dirac Equation from Poincaré Algebra? For the case of Galilean group, the time translation is given by the generator $H$. Hence,
$$\mid\psi(t)\rangle\to \mid\psi(t+s)\rangle =e^{-iHs}\mid\psi(t)\rangle$$ 
Which immediately is the Schrödinger equation,
$$\dfrac{d}{dt}\mid\psi(t)\rangle=-iH\mid\psi(t)\rangle$$
How to arrive at the Dirac Equation from the Poincaré Algebra/Group? I know it will be much more involved. I am not expecting answers that say how Dirac thought of it or trial and error nonsense.
The necessity for Dirac equation should be visible from the Poincaré Algebra/Group?. How does spin come into the picture?
 A: First, note that on the Hilbert space of physical states is realized the projective representations of the Poincare group, precisely, the representations "up to the sign". This can be incorporated automatically by enlarging the Poincare group up to its universal covering group, which is $ISL(2,C)$. 
Here is the general statement which is the point: one-particle representations of the $ISL(2,C)$ with non-zero mass $m$ and a spin $s$ are realized by the spinor tensors $
\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}, \ \frac{m+n}{2}=s$ satisfying the equations
$$
\tag 1
\begin{cases}(\partial^{2}+m^{2})\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} = 0, \\ \partial^{a\dot{b}}\psi_{aa_{1}...a_{n-1}\dot{b}\dot{b}_{1}...\dot{b}_{m-1}}= 0, \quad \partial^{a\dot{b}} \equiv \partial^{\mu}\sigma_{\mu}^{a\dot{b}}, \ \ \text{for } A,B> 1  \end{cases}
$$ 
As the reminder, the undotted indices correspond to the $SL(2,C)$ group (which is the universal covering group for the Lorentz group $SO(3,1)_{\uparrow}$) transformation $N$, while the dotted indices correspond to the complex conjugated transformation $N^{*}$. Finally, $\sigma_{\mu} \equiv (1, \sigma)$, with $\sigma$ being the three Pauli matrices.
You should understand $(1)$ as the equations defining $\psi$ as the function corresponding to the representation on which two Poincare's group Casimir's operators $\hat{P}^{2}$ (translation/4-momentum squared) and $\hat{W}^{2}$ (Pauli-Lubanski squared) have the values $m^{2}$ and $-m^{2}s(s+1)$ correspondingly. This can be shown by expressing these Casimir's invariants in terms of the spinor tensors objects.
Let's denote the tensor $\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}$ by the representation $\left( \frac{n}{2},\frac{m}{2}\right)$ of the $SL(2,C)$ group to which it corresponds. For the spin one half, from the previous statements we have that both of the representations $\left( \frac{1}{2},0\right)$ and $\left( 0, \frac{1}{2}\right)$ are valid; they're given by the two-component spinors. But corresponding equations, which are simply the Klein-Gordon equations, are far from the Dirac equation... So what's the another point?
In fact, neither $\left( \frac{1}{2}, 0\right)$ nor $\left(0,\frac{1}{2} \right)$ describes the theory which is invariant under $P-,T-,C-$ discrete symmetries separately; from this in particular follows that it's impossible to construct the EM interaction theory of these representations with the static interaction law coinciding with the Coulomb law. 
In order to get invariant theory, we need to take the direct sum
$$
\tag 2 \left( \frac{1}{2},0\right) \oplus \left( 0,\frac{1}{2}\right)
$$
of these representations. 
But this direct sum is of course not irreducible representation. In order to make it irreducible (and simultaneously $P-,T-,C-$ invariant), we need to construct a poincare- and $P-,T-,C-$covariant operator which allocates the irreducible representation when acting on the direct sum $(2)$. It's not hard to construct such operator (assuming you can derive $(1)$), and it turns out that it coincides with the Dirac equation operator.
Similar story is true when we consider photons and gravitons. In fact, the massless representations of the Poincare group are characterized by the helicity value, which is the operator's $\hat{h} = \frac{\hat{\mathbf{W}}\cdot \hat{\mathbf P}}{|\hat{\mathbf P}|}$ value, and physically is the projection of the total angular momentum on the direction of motion. Again, the representations with separate values of helicity (say, 1, or -1) are not $P-,T-,C-$invariant. In order to have invariant theory, it's necessary again to take the direct sum, but now without any projectors. In the case of the photons we'll arrive to Maxwell equations on the strength tensor $F_{\mu\nu}$, while in the case of the gravitons we'll arrive to equations on the linearized source-free general relativity Weyl tensor.
