$N=1$ and $N=2$ supersymmetry for non-relativistic electrons I have been following Fred Cooper's Supersymmetry in Quantum Mechanics and I am currently on pages 63/64 where I have now derived the Pauli-Hamiltonian for a non-relativistic electron in an external field. 
$$H=\frac{1}{2m}(\overrightarrow{p}+\frac{e}{c}\overrightarrow{A})^2+\frac{g}{2}\frac{e\hbar}{2mc}\overrightarrow{B} \cdot \overrightarrow{\sigma}\tag{5.16}$$ 
Where the $\frac{g}{2}\frac{e\hbar}{2mc}\overrightarrow{B} \cdot \overrightarrow{\sigma}$ is the result of us imagining  an electron in a purely magnetic field. Cooper then states that $g=2$ and so gives $N=1$ supersymmetry. 
He then finds the resultant supercharge
$$Q_1=\frac{1}{\sqrt{4m}}(\overrightarrow{p}+\frac{e}{c}\overrightarrow{A}) \cdot \overrightarrow{\sigma}.\tag { 5.17}$$


*

*My first questions is how can I show that his claim that $g=2$ implied $N=1$ supersymmetry, mathematically?


He then states when the magnetic field is perpendicular to the electrons motion we have $N=2$ supersymmetry and the introduction of a complex supercharge. 


*Again how do I show that the now perpendicular magnetic field gives $N=2$ supersymmetry, mathematically?

 A: Both answers are answered explicitly in Cooper, Khare, and Sukhatme,"Supersymmetry and quantum mechanics." Physics Reports 251.5-6 (1995): 267-385, eqns (402-404), without excessive nomenclature stretches contrasting N=1 & 2. But since you wish to stick to their book, just heed their exhortation to prove your (5.18), $H=2 Q_1^2$. 
Just resolve the product of the Pauli matrices, as you know it, by inspection,
$$
 \frac{2}{4m}\left((\overrightarrow{p}+\frac{e}{c}\overrightarrow{A}) \cdot \overrightarrow{\sigma}\right)^2=\frac{2}{4m} \left ( (\vec{p}+e\vec{A}/c)\cdot(\vec{p}+e\vec{A}/c) + \frac{ie\hbar}{ic} (\vec{\nabla}\times\vec{A}) \cdot \vec{\sigma} ) \right )=H ~,
$$ 
where the r.h.side is strictly your Hamiltonian (5.16) with g =2, not any g.
Consequently, $\{Q_1,Q_1\}=H$, supersymmetry, and, a fortiori, a symmetry of the hamiltonian,  $[Q_1,\{Q_1,Q_1\} ]=0$, I hope you can see by inspection. 
The second part of the exercise, (5.20-5.24) is also straightforward, something you can do following this, always utilizing the standard Pauli matrix feature. Alert, however: In a breathtakingly unfortunate notation, on the Landau orbit plane, your $Q_1$ here has morphed to that book's $Q^2$ of (5.23), where the "2" is now a new superscript index supplanting the subscript 1, and not a power! Manipulating the Pauli matrix commutators, nevertheless, is trivial.
For example, for the off-diagonal piece of (5.24), defining $\Pi_i\equiv p_i+eA_i/c$, by inspection,
$$
2\{ Q^1,Q^2\}= \{ (-\Pi_y \sigma _x + \Pi_x \sigma_y ) ,(\Pi_x \sigma_x+\Pi_y \sigma_y)  \}\\ = i\sigma_z (-\Pi_y^2 + \Pi_x^2 +\Pi_y^2 - \Pi_x^2) + 1\!\!1 (-\Pi_y \Pi_x -\Pi_x \Pi_y +\Pi_x \Pi_y +\Pi_y \Pi_x  )=0 ~,
$$
etc... It's really straightforward, no?
