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Choose an inscribed regular tetrahedron from one of the regular compounds of five tetrahedra in the dodecahedron, and let $H$ be its rotational stabilizer. Its twelve tetrahedral rotations preserve the dodecahedron. Concretely, with $\varphi =(1+\sqrt{5})/2$, use dodecahedron vertices $(\pm 1,\pm 1,\pm 1)$ together with cyclic coordinate placements of $(0,\pm {\varphi }^{-1},\pm \varphi )$. The tetrahedron has the four cube vertices whose coordinate product is $+1$. Cyclic coordinate permutations and even numbers of coordinate sign changes preserve both vertex sets, have determinant $+1$, and give twelve distinct rotations. No other rotation can preserve this tetrahedron, because a regular tetrahedron has only twelve orientation-preserving symmetries.

Their action on its four vertices is faithful and consists exactly of the even permutations: the eight rotations through $\pm 120^{\circ }$ induce three-cycles, the three half-turns induce double transpositions, and there is the identity. Thus $H\cong A_4$, independently of part (ii).

Finally a nonidentity element of $H$ has order two or three. In the dodecahedron's full rotation group, all fifteen half-turns about opposite-edge axes form one conjugacy class, and all twenty third-turns about opposite-vertex axes form one conjugacy class. These geometric conjugacies follow by carrying the corresponding oriented axes to one another by rotations. A normal subgroup containing such an element must contain its whole class, which already has more than twelve elements. Hence no nonidentity normal subgroup of $G$ is contained in $H$, verifying the required extra property.

Choose one regular tetrahedron inscribed in the dodecahedron and let $H$ be its stabilizer among rotations of the dodecahedron. The twelve rotations of that tetrahedron preserve the dodecahedron; geometrically these are its identity, eight rotations by $\pm 2\pi /3$ about tetrahedral vertex axes and three half-turns about opposite tetrahedral edge axes. They act faithfully on its four vertices and give the twelve even permutations. Thus $\boxed{H\cong A_4,|H|=12}$ without using part (ii).

To check the core condition geometrically, the fifteen half-turns of the dodecahedron, about opposite-edge-midpoint axes, are conjugate under its rotations. A normal subgroup contained in $H$ cannot contain even one of these half-turns, because normality would then force all fifteen into a group of order twelve. On the other hand every nontrivial normal subgroup of $A_4$ contains a half-turn: if it contains a three-cycle, its conjugates and their inverses generate $A_4$; otherwise its nonidentity elements are among the three double transpositions, which are conjugate in $A_4$. Apply this observation to any normal subgroup of the full rotation group lying inside $H$. It must be trivial, proving exactly the required condition.

Replacing row one by $R_1-R_2-R_3-R_4$ does not change the determinant: adding a multiple of another row preserves a determinant. It produces first row $(-2,0,0,0)$, whose cofactor minor is the three-dimensional identity matrix. Thus the determinant is $-2$.

The columns are homogeneous coordinate vectors $(1,v_i{)}^{\top }$ of the four vertices. Subtracting the first column from each other column preserves the determinant and reduces its magnitude to the absolute determinant of the three edge vectors based at $v_1$. That absolute determinant is the parallelepiped volume. A tetrahedron occupies one sixth of that parallelepiped, so $V=|-2|/3!=1/3$ in cubic coordinate units.

A tetrahedron whose six edges have equal length; its four faces are equilateral triangles.

An $n$-simplex is the convex hull of $n+1$ affinely independent points. Every point has a unique expression $x={\sum }_{i=0}^n{\lambda }_i{v}_i$ with ${\lambda }_i\ge 0$ and ${\sum }_i{\lambda }_i=1$, its barycentric coordinates. A zero-simplex is a point, a one-simplex a line segment, a two-simplex a filled triangle, and a three-simplex a tetrahedron. Its faces are obtained by taking subsets of the vertices. This uniform description allows simplicial complexes to build spaces from elementary pieces in every dimension.

For a positively oriented basis $\mathbf{a},\mathbf{b},\mathbf{c}$, put $\tau =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The center of an interior sphere at distance $r$ from each of the three tetrahedron face planes through the origin has the displayed expression. The three inward unit vectors are the normalized cross products $\mathbf{b}\times \mathbf{c}$, $\mathbf{c}\times \mathbf{a}$, $\mathbf{a}\times \mathbf{b}$. Taking inner products with these vectors solves three independent signed-distance equations. Reversing some signs instead produces centers on exterior sides. Existence of a sphere contained in the full tetrahedron also requires clearance from the fourth face.

By Melvin A. Neuman, Trinh-Toan, Lawrence F. Dahl.

Journal of the American Chemical Society 94, 3383–3388. Original diffraction study: four bonded Fe vertices form a tetrahedron; four triply bridging carbonyl carbons form an interpenetrating tetrahedron.

The 24-operation molecular point group of an ideal regular tetrahedron, including improper operations. Its irreducible representations are $A_1,A_2,E,T_1,T_2$, with dimensions $1,1,2,3,3$. The polar-vector representation is $T_2$, the axial-vector representation is $T_1$, and symmetric quadratic functions transform as $A_1+E+T_2$.

Four nearest neighbours at tetrahedron vertices around a central site.

A position surrounded by four nearest sites at tetrahedron vertices. A conventional face-centred cubic cell has eight such sites, at all combinations of quarter and three-quarter fractional coordinates. Whether occupancy constitutes a defect depends on the reference crystal.

Take a central chemical atom at the origin and four substituents at the vertices of an ideal regular tetrahedron:These dimensionless coordinates describe shape, not measured bond lengths. For distinct labels A, B, C and D, define a signed determinantIt is six times the oriented volume of the labelled tetrahedron. A translation cancels from the differences. A rotation $Q$ has $\det Q=+1$ and preserves $\Delta$, whereas the mirror matrix $S=\mathrm{diag}(-1,1,1)$ has $\det S=-1$ and changes it to $-16$. Consequently no rotation can place every distinct label on its mirror partner. This argument uses three-dimensional coordinates, not clockwise order on a page.

Now let C and D both be the same substituent H. Swapping these identical groups is allowed. For this model the half-turnabout the axis $(0,1,1)$ maps the reflected A and B to their original positions and exchanges the two H positions. The models superpose. With distinct C and D, the same half-turn fails precisely at those two labels; the determinant argument rules out every other rotation as well. This is the local reason two identical hydrogen chemical atoms prevent glycine's amino acid alpha carbon from being a tetrahedral stereocenter.

For a flexible molecule, also distinguish conformational isomerism from a persistent enantiomer pair. A particular frozen shape may be chiral while ordinary internal motions interconvert mirror shapes. Turning the entire model is always allowed in a spatial comparison; an internal motion must be considered with its actual accessibility and timescale.

In tetrahedral molecular geometry, four neighbouring chemical atoms surround a central chemical atom in the directions of the vertices of a tetrahedron. In the regular idealization, the central atomic nucleus lies at the centre of a regular tetrahedron, and every pair of chemical bonds makes the same bond angle. Methane, ${\mathrm{CH}}_4$, provides the standard example: its carbon has four bonding electron domains and no lone pair. See Chemistry 2e, §7.6: tetrahedral methane.

The exact ideal bond angle follows from four directions proportional to $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$, and $(-1,-1,1)$. Each has squared Euclidean norm $3$, and each distinct pair has dot product $-1$. HenceThis is a statement about a regular tetrahedron, rather than a universal exact value for every molecule labelled tetrahedral. It also explains why a cross-shaped flat drawing with $90^{\circ }$ gaps is misleading: four equal directions can spread out more evenly in three dimensions.

The same vectors sum to zero. Consequently, four equal bond-dipole contributions along those directions cancel. The symmetric methane molecule is nonpolar, but a tetrahedral shape alone does not imply a nonpolar molecule: replacing one neighbour can make the electronic distribution and bond contributions unequal. Chemical polarity depends on both molecular geometry and electron density. See McMurry, Organic Chemistry, §2.2: molecular dipoles and methane.

Four equivalent $s$ atomic orbitals at the vertices of a regular tetrahedron carry the tetrahedral symmetry representation $A_1\oplus T_2$. The normalized equal sum is $A_1$; three coordinate-sign combinations span $T_2$. Every normalized coefficient vector has entries $\pm 1/2$ in a suitable orthonormal real basis. These functions simplify a Hamiltonian invariant under the tetrahedral operations, but equivalent symmetry copies on other sites may still mix.

Place the centroid of a tetrahedron at the origin, so its four vertex vectors satisfy $r_1+r_2+r_3+r_4=0$. Midpoints of opposite edges have opposite vectors, hence equal distances from the centroid. For the three opposite-edge pairs, choose midpoint vectors $(r_1+r_j)/2$, $j=2,3,4$. Expanding their squared lengths and using ${\sum }_{j=2}^4{r}_j=-r_1$ givesThe argument uses no regularity or equal-edge assumption on the tetrahedron.

Let $D=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The reciprocal basis gives the coordinates of any position vector asThe point lies in the closed tetrahedron with vertices $0,\mathbf{a},\mathbf{b},\mathbf{c}$ exactly when $\alpha ,\beta ,\gamma \ge 0$ and $\alpha +\beta +\gamma \le 1$. Indeed, these conditions express it as a convex combination of the four vertices, with fourth weight $1-\alpha -\beta -\gamma$. They also describe the intersection of the four inward half-spaces bounded by its faces. Strict inequalities characterize its interior.

Place one pair of opposite tetrahedron vertices on the line $b+\lambda a$ and the other pair on $d+\mu c$, with fixed nonzero direction vectors $a,c$. Let $\Delta =d-b$ and let $\delta \lambda ,\delta \mu$ be the parameter differences within the two pairs. The scalar triple product of the three edge vectors from one vertex gives the displayed volume. Terms parallel to $a$ or $c$ vanish in the triple product, leaving only $\Delta \cdot (a\times c)$. Sliding either pair along its own line while preserving its separation leaves this volume unchanged. When the two lines are skew lines, it can also be written $V=d_{\perp }{\ell }_1{\ell }_2|\sin \alpha |/6$, where $d_{\perp }$ is their shortest distance, ${\ell }_1,{\ell }_2$ the two edge lengths and $\alpha$ their direction angle. Intersecting or parallel lines give zero volume because all four vertices are coplanar.

The three-uniform tetrahedron is the hypergraph on four vertices with all four possible triple hyperedges. It is the complete uniform hypergraph $K_4^3$, and differs from its geometric realization as a tetrahedron. Its edge pattern encodes four-term arithmetic progressions.

Trinh-Toan helped characterize an iron carbonyl tetramer by diffraction. Its four metal atoms form a bonded tetrahedron, interpenetrated by a second tetrahedron of triply bridging carbonyl carbon atoms. The result is a concrete structural example of a metal metal cluster.^[ref]