A screw axis consists of two symmetry operations performed in tandem: rotation and translation. Specifically, an $n_{m}$ screw axis is defined as a rotation of $\tfrac{2\pi}{n}$ radians about an $n$-fold axis followed by a translation of $\tfrac{m}{n}$ units in a direction parallel to that $n$-fold axis. As an illustrative example, consider a $2_{1}$ screw axis parallel to the unit cell vector b (i.e., space group $\text{P}2_{1}$). To apply a $2_{1}$ screw axis, we would rotate by $\frac{2\pi}{2}=\pi$ radians and translate $\tfrac{1}{2}$ units along b. In terms of transformation matrices, it's straightforward to express this as a rotation followed by a translation: $$\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} + \begin{bmatrix} 0 \\ \tfrac{1}{2} \\ 0 \end{bmatrix} = \begin{bmatrix} -x \\ \tfrac{1}{2} + y \\ -z \end{bmatrix} \\$$ This expression is essentially telling us that a $2_{1}$ screw axis operates on every atom at coordinates $(x, y, z)$ to generate an analogous symmetry-equivalent atom at coordinates $(-x, \tfrac{1}{2} + y, -z)$. Now recall that the structure factor $F_{hkl}$ is obtained by summation over all atoms within the unit cell: $$F_{hkl}=\sum_{j} f_{j}(s)\,\mathrm{exp}[2\pi i\,(hx_{j} + ky_{j} + lz_{j})],$$ where $f_{j}(s)$ is the individual scattering factor for the $j^{\mathrm{th}}$ atom, $h$, $k$, and $l$ correspond to the Miller indices, and $x_{j}$, $y_{j}$, and $z_{j}$ give the fractional coordinates of the $j^{\mathrm{th}}$ atom in real space. Since this equation for $F_{hkl}$ is generic, let's add some specificity by updating it with the symmetry relation we derived for a $2_{1}$ screw axis: $$F_{hkl}=\sum_{j} f_{j}(s)\,\mathrm{exp}[2\pi i\,(hx_{j} + ky_{j} + lz_{j})]+f_{j}(s)\,\mathrm{exp}[2\pi i\,(-hx_{j} + k\{\tfrac{1}{2} + y_{j}\} - lz_{j})]. $$ Now let's focus our analysis specifically on the $(0k0)$ set of reciprocal lattice points. If we substitute $h = 0$ and $l = 0$ and factor out $f_{j}(s)$, we obtain $$F_{0k0}=\sum_{j} f_{j}(s)\,\{\mathrm{exp}[2\pi i\,(ky_{j})]+\mathrm{exp}[2\pi i\,(\tfrac{1}{2}k + ky_{j})]\}.$$ At first glance, this may seem like a dead end. To simplify this expression, however, we can invoke a few mathematical tricks. First, recall that $\mathrm{exp}(\theta + \phi) = \mathrm{exp}(\theta) \cdot \mathrm{exp}(\phi)$. Using this identity, we can deconstruct the second term. Let's start by distributing $2\pi i$, which gives us $$F_{0k0}=\sum_{j} f_{j}(s)\,\{\mathrm{exp}[2\pi i\,(ky_{j})]+\mathrm{exp}[k\pi i + 2\pi i (ky_{j})]\}.$$ Now we expand using $\mathrm{exp}(\theta + \phi) = \mathrm{exp}(\theta) \cdot \mathrm{exp}(\phi)$, which yields $$F_{0k0}=\sum_{j} f_{j}(s)\,\{\mathrm{exp}[2\pi i\,(ky_{j})]+\mathrm{exp}[k\pi i] \cdot \mathrm{exp}[2\pi i (ky_{j})]\}. $$This form makes it clear that we have a common factor of $\mathrm{exp}[2\pi i\,(ky_{j})]$. By factoring this out, we obtain $$F_{0k0}=\sum_{j} f_{j}(s)\,\{\mathrm{exp}[2\pi i\,(ky_{j})] \cdot [1 + \mathrm{exp}(k\pi i)]\}.$$ We still have one more trick up our collective sleeves. It turns out we can express $\mathrm{exp}(k\pi i)$ as a power of $-1$, a property which stems from Euler's identity: $\mathrm{exp}(\pi i) = -1$. Although a rigorous proof of this idea falls outside the ambit of crystallography, it should make intuitive sense that the function $f(k)=(-1)^k$ naturally possesses both real and imaginary components (for instance, what happens when we substitute $k = \tfrac{1}{2}$?). Try plotting this if you can't visualize it. Substituting $\mathrm{exp}(k\pi i)=(-1)^k$, we find $$F_{0k0}=\sum_{j} f_{j}(s)\,\mathrm{exp}[2\pi i\,(ky_{j})] \cdot [1 + (-1)^{k}].$$ Since the second term is just an innocent coefficient, we can remove it from the summation: $$ F_{0k0}=[1 + (-1)^{k}] \sum_{j} f_{j}(s)\,\mathrm{exp}[2\pi i\,(ky_{j})].$$ We're almost there. In this form, it's facile to analyze what happens when we supply certain integer arguments for $k$ (remember that the Miller indices $h$, $k$, and $l$ are integers by definition, so we can safely disregard non-integer arguments). When $k$ is odd, we can see that $[1 + (-1)^{k}]$ evaluates to 0, leading to $F_{0k0}=0$. When $k$ is even, however, $[1 + (-1)^{k}]$ evaluates to 2, producing a nonzero value for $F_{0k0}$. (In other words, for any integer input, the function $f(k) = 1 + (-1)^{k}$ oscillates between outputs of 0 and 2.) We can express this formulaically as $$ \boxed{F_{0k0}=[1 + (-1)^{k}] \sum_{j} f_{j}(s)\,\mathrm{exp}[2\pi i\,(ky_{j})]\begin{cases} =0,& \text{if } k=2n+1\\ \neq 0, & \text{if } k=2n \end{cases}}$$ where $n$ is some integer (this $n$ notation is just a formalistic way of expressing $k$ as either even or odd). So for every putative Bragg peak $(0k0)$ where $k$ is odd, we should expect to observe nothingness (i.e., systematically extinct reflections). Let's say we index our diffraction dataset and notice zero intensities for reflections $(030)$, $(050)$, or $(070)$. However, we did successfully record strong intensities for reflections $(020)$, $(040)$, $(060)$, etc. This particular pattern of systematic absences would provide robust evidence that our crystal possesses a $2_{1}$ screw axis along b. Analogously, we can derive similar selection rules for $2_{1}$ screw axes parallel to a and c. These turn out to yield systematic absences for $(h00)$ and $(00l)$ reflections when $h$ or $l$ is odd, respectively. One of the most frequently observed space groups¹ for protein crystals is $\text{P}2_{1}2_{1}2_{1}$, which features $2_{1}$ screw axes along all three unit cell vectors. A comprehensive list of extinction conditions for every crystallographically permissible screw axis is given below.

$$ \begin{array}{|c|c|c|} \hline \begin{array}{c} \text { Zone-axis reflections } \end{array} & \begin{array}{c} \text { Extinction conditions } \end{array} & \begin{array}{c} \text { Screw axis } \end{array} \\ \hline (h00) & h\neq2 n & 2_{1} \parallel \textbf{a} \\ \hline (h00) & h\neq4 n & 4_{1}, 4_{3} \parallel \textbf{a} \\ \hline (0k0) & k\neq2 n & 2_{1} \parallel \textbf{b} \\ \hline (0k0) & k\neq4 n & 4_{1}, 4_{3} \parallel \textbf{b} \\ \hline (00l) & l\neq2 n & 2_{1}, 4_{2}, 6_{3} \parallel \textbf{c} \\ \hline (00l) & l\neq3 n & 3_{1}, 3_{2}, 6_{2}, 6_{4} \parallel \textbf{c} \\ \hline (00l) & l\neq4 n & 4_{1}, 4_{3} \parallel \textbf{c} \\ \hline (00l) & l\neq6 n & 6_{1}, 6_{5} \, \parallel \, \textbf{c} \\ \hline \end{array}$$

Crucially, note that systematic absences in reciprocal space always correspond to the translational component of the symmetry operator in real space (i.e., $\textbf{a} \longleftrightarrow h$, $\textbf{b} \longleftrightarrow k$, $\textbf{c} \longleftrightarrow l$). More generally, a similar argument can be articulated for any screw axis, glide plane, or nonprimitive lattice—in other words, any element of symmetry which carries a translational component. A glide plane, for instance, also consists of two symmetry operations performed in tandem: reflection and translation. The nomenclature for a glide plane simply denotes the axis to which it's parallel (i.e., $a$, $b$, and $c$ for unit cell vectors, as well as $n$ and $d$ for more complicated diagonals). For clarity's sake, let's cover the simplest space group which contains a glide plane: Pc, which features a $c$-glide perpendicular to the unit cell vector $\textbf{b}$. This indicates that the $y$-coordinate is inverted by the reflection operation, which is subsequently followed by a translation parallel to $\textbf{c}$: \begin{equation} \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ \tfrac{1}{2} \end{bmatrix} = \begin{bmatrix} x \\ -y \\ \tfrac{1}{2}+z \end{bmatrix} \end{equation} This expression is telling us that a $c$-glide plane perpendicular to the unit cell vector $\textbf{b}$ takes every atom at coordinates $(x, y, z)$ and replicates it at coordinates $(x, -y, \tfrac{1}{2}+z)$. Using the same logic as before, we can now incorporate this symmetry operator into the structure-factor equation: \begin{equation} F_{hkl}=\sum_{j} f_{j}(s)\,\exp[2\pi i\,(hx_{j} + ky_{j} + lz_{j})]+f_{j}(s)\,\exp[2\pi i\,(hx_{j} - ky_{j} + l\{\tfrac{1}{2} + z_{j}\})]. \end{equation} The relevant zone axis this time is $(h0l)$. Substituting $k = 0$ and factoring out $f_{j}(s)$, we obtain \begin{equation} F_{h0l}=\sum_{j} f_{j}(s)\,\{\exp[2\pi i\,(hx_{j} + lz_{j})]+\exp[2\pi i\,(hx_{j} + \tfrac{1}{2}l + lz_{j})]\}. \end{equation} Fast-forwarding through a few very similar simplification steps, we ultimately arrive at a familiar-looking result: \begin{equation} \boxed{F_{h0l}=[1 + (-1)^{l}] \sum_{j} f_{j}(s)\,\exp[2\pi i\,(hx_{j} + lz_{j})]\begin{cases} =0,& \text{if } l=2n+1\\ \neq 0, & \text{if } l=2n \end{cases}} \end{equation} A list of extinction conditions for commonly observed glide planes is given below. Unlike the corresponding table for screw axes, this compilation is not comprehensive—it only includes axial glide planes perpendicular to one of the three unit cell vectors (some more exotic achiral space groups feature so-called oblique glides perpendicular to specific low-index Bragg planes such as [120] or [110], which I have not derived here).

$$ \begin{array}{|c|c|c|} \hline \begin{array}{c} \text { Zone-axis reflections } \end{array} & \begin{array}{c} \text { Extinction conditions } \end{array} & \begin{array}{c} \text { Glide plane } \end{array} \\ \hline (0kl) & k\neq2 n & b \perp \textbf{a} \\ \hline (0kl) & l\neq2 n & c \perp \textbf{a} \\ \hline (0kl) & k+l \neq2 n & n \perp \textbf{a} \\ \hline (0kl) & k+l \neq4 n & d \perp \textbf{a} \\ \hline (h0l) & l\neq2 n & c \perp \textbf{b} \\ \hline (hk0) & h\neq2 n & a \perp \textbf{c} \\ \hline \end{array}$$

In sum, the central idea we've explored here—i.e., that certain symmetry elements in real space always leave indelible footprints in reciprocal space—is a cornerstone of crystallography. Hunting for systematic absences is one of the most crucial steps in deducing the correct space group to which an unknown crystal rightfully belongs.