Bravais lattice - Why is every reflex allowed in Bravais lattice? I cannot wrap my head around why every reflex in a Bravais lattice (3-Dimensional) is allowed, according to my book, it's because there is only one term in the expression for the structure factor, but I still don't understand.
I must've missed something important involving Bravais lattice but I cannot wrap my head around what that would be. 
 A: I think maybe you have misunderstood what every reflex is allowed means. In every 3-Dimensional Lattice reflexes are possible if the von Laue condition
$$\vec{k'} = \vec{k} + \vec{G}$$
is satisfied. Here $\vec{G}$ denotes a vector in the reciprocal lattice and $\vec{k}$ and $\vec{k'}$ are the incoming and reflected wave vectors.
Now on to the structure factor. The scatered wave is given by
$$\psi(t) = e^{-i\omega t}\int\rho(r)e^{i (k-k')}d^3r$$
where $\rho$ denotes the scattering density. We can write this scattering density as a convolution of the Lattice with the scattering density inside a primitive cell. The convolution of these two is then basically a repetition of the unit cell scattering density in the same manner as the lattice points. 
We obtain 
$$\psi \propto N \int\rho_b(r)e^{-ikG}d^3r = N * S_G$$
where $\rho_b$ is the scattering density inside a unit cell. If we calculate S for a primitive cubic lattice with two different Basis Atoms for example CsCl with a Caesium at $(000)$ and a Cl at ($\frac{1}{2}\frac{1}{2}\frac{1}{2})$ we get
$$S_{hkl} = f_1 + f_2 e^{-i\pi(h+k+l)} = \left\{\begin{array}{lr}
        f_1 + f_2, & \text{for } h+k+l \ even\\
        f_1 - f_2, & \text{for } h+k+l \ odd\\
        \end{array}\right\}$$
So we see that the reflexes corresponding to some reciprocal lattice vectors are suppressed if $f_1$ and $f_2$ are close to each other.
This happens because whe have a multi atom base. In the simple bravais lattice this cannot happen however and thats what the author of your text had in mind.
