Derivation of ‘Unknown’ Symmetry Elements 2
The \(n\) glide perpendicular to \(a\) converts \((x, y, z)\) to \((-x, y+\frac{1}{2}, z+\frac{1}{2})\).
If the mirror plane is applied next, this inverts the \(b\) coordinate, so we arrive at \((-x, -y + \frac{1}{2}, z+\frac{1}{2})\). Remember that with fractional coordinates and periodic boundary conditions that \((-x, -y + \frac{1}{2}, z+\frac{1}{2})\) and \((-x, -y – \frac{1}{2}, z+\frac{1}{2})\) are identical. This new position is equivalent to applying a \(2_{1}\) screw axis parallel to \(c\) on our original position.
If the \(a\) glide is applied next, this inverts the \(c\) coordinate and adds \(\frac{1}{2}\) to the \(a\) coordinate. Thus, applying the \(n\) glide followed by the \(a\) glide, we arrive at the position \((-x+\frac{1}{2}, y + \frac{1}{2}, -z + \frac{1}{2})\). This is equivalent to the application of a \(2_{1}\) screw axis parallel to \(b\).
Relative Symmetry Element Positions
The method above assumes that the symmetry elements you are applying are at the origin. As the origin can be placed anywhere, this is not an issue. It should be noted that symmetry elements in typical space group diagrams do not always operate at the origin. This method informs about the relative position of different symmetry elements. Consider the application of the \(n\) glide and \(a\) glide above. The final position arrived at was \((-x+\frac{1}{2}, y + \frac{1}{2}, -z + \frac{1}{2})\). This is equivalent to the application of a \(2_{1}\) screw axis parallel to \(b\), located at \((\frac{1}{4}, b, \frac{1}{4})\). This is the position of the \(2_{1}\) screw axis if the \(n\) glide and \(a\) glide pass through the origin. In the actual space group diagram in The International Tables for Crystallography, the origin may be placed in a different position.
Inversion Centres
Inversion centres are not listed in orthorhombic space groups as they can be derived from other symmetry elements.
Any orthorhombic space group with mirror or glide planes perpendicular to all three axes will have an inversion centre present. Each mirror or glide plane will invert each axis coordinates, arriving at the position \((-x, -y, -z)\), depending on where the origin is placed. Therefore an inversion centre must be present. Feel free to give this a go with a space group of your choosing (or \(Cccm\), if you need one given).