Q:

How to construct a supersymmetry algebra?

Starting with the general notion of supersymmetry:
$$Q| boson \rangle = | fermion \rangle  \\ 
Q| fermion \rangle = | boson \rangle$$
I want to construct the superalgebra relations. After applying $Q$ twice we get:
$$
Q^2| boson \rangle = | boson \rangle; \\ Q^2| fermion \rangle = | fermion \rangle
$$ 
Now, we can use the identical particle commutation and anti-commutation relations. And since $Q$ is both bosonic and fermionic this must hold:
$$\{Q_\alpha,Q_\beta\} = \left[Q_\alpha,Q_\beta\right] = 0 \\ 
\{Q_\alpha,Q_\beta ^\dagger \} = \left[Q_\alpha,Q_\beta ^\dagger \right] = \delta_{\alpha \beta}$$
Where we assume $Q$ are elements of Poincaré algebra namely:
$$
[Q_\mu, Q_\nu] = 0 \\
-i ~[M_{\mu\nu}, Q_\rho] = \eta_{\mu\rho} Q_\nu - \eta_{\nu\rho} Q_\mu \\ 
-i ~[M_{\mu\nu}, M_{\rho\sigma}] = \eta_{\mu\rho} M_{\nu\sigma} - \eta_{\mu\sigma} M_{\nu\rho} - \eta_{\nu\rho} M_{\mu\sigma} + \eta_{\nu\sigma} M_{\mu\rho}
$$
The combination of being a Poincaré algebra and obeying the fermions and bosons commutations relations should make it a  Super-Poincaré algebra, but I am not sure how to show it.

A:

For me the most satisfying exposition is either Quantum Fields and Strings: A Course for Mathematicians or Wess and Bagger's Supersymmetry and Supergravity. The latter starts off with providing the motivation from the Coleman-Mandula theorem, leading to graded Lie algebras.
In particular, we start off with the general ansatz $\{Q,Q'\} = X, [X,X']= X''$ and $[Q,X] = Q''$. We have that $X$ can only be elements of the Poincaré algebra or elements of a Lorentz invariant compact Lie algebra $\mathscr A = \mathscr A_1 \oplus \mathscr A_2$ where $\mathscr A_1$ is semisimple and $\mathscr A_2$ abelian.
On the other hand, the former book is a masterful approach, starting with a great degree of generality and rigour, starting off with the properties of $\mathbb Z_2$ graded objects, such as vector spaces and manifolds.

