Deriving the 14 Bravais Lattices
Select any of the 17 plane groups below. Watch how symmetry elements (like mirrors and rotation axes) form infinite 3D boundaries. To prevent breaking these symmetries, the extrusion vector ($c$) must snap to specific alignments, revealing the 14 Bravais Lattices.
The 17 Plane Groups
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Comprehensive Mapping of All 17 Plane Groups
Detailed breakdown mapping every single 2D plane group to its permitted 3D extrusion conditions, resulting crystal systems, and specific Bravais lattices.
Why can't pm, pg, or cm form an Orthorhombic Lattice?
In 3D, the defining symmetry of the Orthorhombic system is point group $222$, $mm2$, or $mmm$. This requires intersecting symmetry elements (like two orthogonal mirrors, or a mirror and a perpendicular 2-fold axis).
Plane groups like pm, pg, and cm contain only a single direction of symmetry planes (parallel mirrors or parallel glides). The only constraint these single planes impose on the extrusion vector ($c$) is that it must lie parallel to the plane. Because $c$ is free to tilt within that plane, it does not force an orthogonal $90^\circ$ angle. Therefore, the general symmetry it enforces is strictly Monoclinic.
While you could manually pull the vector perfectly straight up ($\alpha=\beta=\gamma=90^\circ$), this is merely an "accidental metric symmetry" (a pseudo-orthorhombic monoclinic cell). It does not upgrade the actual internal crystal symmetry to Orthorhombic because it still lacks that crucial second intersecting symmetry plane. To truly force an Orthorhombic Bravais lattice, you must start with a 2D plane group that already has orthogonal constraints, such as pmm, pmg, pgg, or cmm.
Summary of 3D Crystallography
| Crystal System | Unit Cell Properties | Basic Symmetry | Underlying Plane Group | Point Groups (Schoenflies) | Point Groups (Hermann-Mauguin) | Space Groups |
|---|---|---|---|---|---|---|
| Triclinic (P) | $a \neq b \neq c$ $\alpha \neq \beta \neq \gamma$ |
No symmetry | p1 | $C_1, C_i$ | $1, \bar{1}$ | 2 |
| Monoclinic (P, A/B/C or I) |
$a \neq b \neq c$ $\alpha = \beta = 90^\circ, \gamma \neq 90^\circ$ |
2 or m | p2, pm, pg, cm | $C_2, C_s, C_{2h}$ | $2, m, 2/m$ | 13 |
| Orthorhombic (P, A/B/C, I, F) |
$a \neq b \neq c$ $\alpha = \beta = \gamma = 90^\circ$ |
2mm | pmm, pgg, pmg, cmm | $C_{2v}, D_2, D_{2h}$ | $mm2, 222, mmm$ | 59 |
| Tetragonal (P, I) |
$a = b \neq c$ $\alpha = \beta = \gamma = 90^\circ$ |
4 or 4mm | p4, p4m, p4g | $C_4, S_4, C_{4h}, C_{4v}, D_{2d}, D_4, D_{4h}$ | $4, \bar{4}, 4/m, 4mm, \bar{4}2m, 422, 4/mmm$ | 68 |
| Rhombohedral (R) | $a = b = c$ $\alpha = \beta = \gamma \neq 90^\circ$ |
3 | p3, p3m1 | $C_3, S_6, D_3, C_{3v}, D_{3d}$ | $3, \bar{3}, 32, 3m, \bar{3}m$ | 25 |
| Hexagonal (P) | $a = b \neq c$ $\alpha = \beta = 90^\circ, \gamma = 120^\circ$ |
6 and 3 | p3, p3m1, p31m, p6, p6m | $C_6, C_{3h}, C_{6v}, C_{6h}, D_6, D_{3h}, D_{6h}$ | $6, \bar{6}, 6mm, 6/m, 622, \bar{6}m2, 6/mmm$ | 27 |
| Cubic (P, I, F) |
$a = b = c$ $\alpha = \beta = \gamma = 90^\circ$ |
4 and 3 or 2 and 3 | p4, p4m, p4g | $O_h, O, T_d, T_h, T$ | $m\bar{3}m, 432, \bar{4}3m, m\bar{3}, 23$ | 36 |
Did It Sink In? - 10-11
Test your mastery of 3D lattice derivation, plane group extrusion constraints, and historical crystallographic trivia.