The two-face corner group — PGL₂(𝔽₅) ≅ S₅, the miracle on six points

Take the subgroup of the cube generated by two adjacent face turns (say R, U), looking only at corner permutations (ignore twist). The two faces share 6 corners (4 + 4 − 2). Naïvely 6! = 720, but the actual order is 120 = 5!. This 120-element group is ${\mathrm{PGL}}_2({\mathbb{F}}_5)\cong S_5$ in its sharply 3-transitive action on six points; it also gives an exotic transitive embedding $S_5\hookrightarrow S_6$, and that embedding is the source of the unique non-trivial outer automorphism of $S_6$. This is the densest cluster of coincidences in finite group theory.

30.1 The two-face six-corner puzzle

Turn R and U on a full 3×3×3 (or 2×2×2) any number of times and observe the six corners in the shared R-and-U region: which slots do they land in? The question was first catalogued by Jaap Scherphuis (jaapsch.net/puzzles/pgl25.htm). Because we only use these two faces, the six corners never leave their six slots. A naïve upper bound is 6! = 720, but empirically only 120 distinct permutations occur — exactly 5!. This 120-element group admits two equivalent descriptions:

• Combinatorial (30.2): six corners partition into three pairs in 15 ways, falling into 5 essentially different pair patterns V W X Y Z. The group $⟨R,U⟩$ acts faithfully on these 5 patterns and realises all of $S_5$.
• Projective (30.3–30.5): label the six corners by ${\mathbb{P}}^1({\mathbb{F}}_5)=\{0,1,2,3,4,\infty \}$. Each of R, U realises a Möbius transformation (a 2×2 mod-5 matrix of non-zero determinant); together they generate the full ${\mathrm{PGL}}_2({\mathbb{F}}_5)$.

Equivalence (30.2 ↔ 30.3) is a finite Cayley correspondence: in ${\mathrm{PGL}}_2({\mathbb{F}}_5)$, "pairings of 6 points" = "fixed pairs of involutions" = "matrices mod centre" all have dimension 5 (cosets of the Klein 4-group). This 5 is exactly why 120 of the 720 permutations remain.

30.2 Pair-pattern proof

Six corners split into three pairs in $\left(\genfrac{}{}{0ex}{}{6}{2}\right)\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{2}{2}\right)/3!=15$ ways. But we want R-and-U-stable pair sets; the group $⟨R,U⟩$ groups the 15 pairs into 5 equivalence classes of 3 each, labelled V W X Y Z. Each face turn permutes the 5 labels by some element of $S_5$.

The arithmetic $5=15/3$ uses a non-trivial fact: each of the 5 equivalence classes contains exactly 3 pairs. This reflects centraliser structure in ${\mathrm{PGL}}_2({\mathbb{F}}_5)$ — each involution's fixed-pair set is a pair-pattern, and the involutions fall into 5 conjugacy classes × 3 involutions each = 15. More in 30.3.

30.3 ℙ¹(𝔽₅) and PGL₂(𝔽₅)

In affine coordinates, each $[x:y]$ with $y\ne 0$ is $[z:1]$ for $z=x/y\in F$, plus a single point at infinity $[1:0]$ denoted $\infty$. Over ${\mathbb{F}}_5$:$${\mathbb{P}}^1({\mathbb{F}}_5)=\{0,1,2,3,4,\infty \},|{\mathbb{P}}^1({\mathbb{F}}_5)|=6.$$

Order (general $q$): ${\mathrm{GL}}_2({\mathbb{F}}_q)$ = "two linearly independent columns" = $(q^2-1)(q^2-q)$; $|Z|=q-1$. So

For $q=5$: $|{\mathrm{PGL}}_2({\mathbb{F}}_5)|=5\cdot 6\cdot 4=120=5!$, the same as $|S_5|$. They are in fact isomorphic; 30.4 provides the explicit witness.

Small table (sanity check):

The coincidence ${\mathrm{PGL}}_2\cong S_n$ happens only for $q\in \{2,3,5\}$ (i.e. $n=q+1\in \{3,4,6\}$). $q=5$ is the final and most dramatic member. Moreover ${\mathrm{PSL}}_2({\mathbb{F}}_5)\cong A_5$ is the smallest non-Abelian simple group (60 elements) and the rotation group of the icosahedron (30.9).

30.4 The sharply 3-transitive action

Proof sketch (existence): we show any triple $(a,b,c)$ of distinct points can be moved to the standard triple $(0,1,\infty )$; composition then takes any to any. Explicitly:

satisfies $f(a)=0,f(b)=1,f(c)=\infty$. (Verify by direct substitution. If any of $a,b,c$ equals $\infty$, replace $z$ by $1/(z-a)$ first; the formula stays valid in the limit.)

Proof sketch (uniqueness): if two Möbius transformations $f_1,f_2$ both send $(a,b,c)\to (a^{\prime },b^{\prime },c^{\prime })$, then $f_2^{-1}\circ f_1$ fixes three points. But a Möbius $z\mapsto (\alpha z+\beta )/(\gamma z+\delta )$ with three fixed points must satisfy $\alpha z+\beta =z(\gamma z+\delta )$ at three values of $z$; a degree-≤2 polynomial with three roots must be zero, forcing $\gamma =0,\alpha =\delta ,\beta =0$ — the identity. ∎

Count check: sharp 3-transitivity gives $|G|=6\cdot 5\cdot 4=120=|{\mathrm{PGL}}_2({\mathbb{F}}_5)|$. ✓

30.5 The cross-ratio — the only invariant

Three points can be moved freely; four points must have one invariant — otherwise sharp 3-transitivity would silently upgrade to 4-transitivity, but $|G|=120\ne 6\cdot 5\cdot 4\cdot 3$. The invariant is the cross-ratio:

Sketch: by 30.4, $(a,b;c,d)=f_{a,b,c}(d)$. For any $\varphi$, uniqueness in 30.4 forces $f_{\varphi (a),\varphi (b),\varphi (c)}=f_{a,b,c}\circ {\varphi }^{-1}$, so $f_{\varphi (a),\varphi (b),\varphi (c)}(\varphi (d))=f_{a,b,c}(d)$. ∎

Possible values: over ${\mathbb{F}}_5$, a cross-ratio of 4 distinct points lies in ${\mathbb{P}}^1({\mathbb{F}}_5)\setminus \{0,1,\infty \}=\{2,3,4\}$ (the values 0, 1, ∞ correspond to degenerate quadruples). So just 3 values up to ordering of the 4 points (and the Klein-4 action by point-swaps preserves the cross-ratio). The detailed orbit counting is a nice exercise.

30.6 The outer automorphism of S₆ — the unique exception

Two embeddings $S_5\hookrightarrow S_6$:

Since $\mathrm{Aut}(S_5)=S_5$ internally, the two embeddings "should be equivalent" inside $S_5$ — but their images in $S_6$ are not conjugate (one is transitive, the other not). This pair of non-conjugate copies of $S_5$ inside $S_6$ is the heart of the outer automorphism. Formally: a bijection sending the exotic $S_5$ to the standard $S_5$ (via 6 synthematic totals ↔ original 6 points, see 30.7) extends to a non-inner $\varphi :S_6\to S_6$.

Check: transpositions in $S_6$ number $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$; cycle-type $(2,2,2)$ elements number $\frac{6!}{2^3\cdot 3!}=\frac{720}{48}=15$. ✓

30.7 Synthemes & duads — Sylvester's six totals

Sylvester (1844) gave a purely combinatorial construction of the outer automorphism using duads and synthemes:

• Duad = a 2-subset of {1..6}. Total $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$.
• Syntheme = a partition of {1..6} into 3 duads. Count: $\frac{6!}{2^3\cdot 3!}=15$, so 15 synthemes.
• Synthematic total = a set of 5 synthemes whose 15 duads cover all $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ duads (each duad appearing exactly once). Equivalently: a 1-factorisation of $K_6$ (partition the 15 edges into 5 perfect matchings). There are exactly 6 totals.

30.8 Mathieu connection — from PGL₂(𝔽₅) to M₂₄

The 120-element ${\mathrm{PGL}}_2({\mathbb{F}}_5)$ sits at the foot of a tower of multiply transitive extensions. Mathieu's five sporadic simple groups $M_{11},M_{12},M_{22},M_{23},M_{24}$ (1861, 1873) form this tower:

$M_{12}$ is the largest sharply 5-transitive group (Jordan, 1872: no sharp $k$-transitive group beyond $S_n,A_n$ exists for $k\ge 6$). $M_{12}$ is the only Mathieu with a non-trivial outer automorphism — built by exactly the same "exotic transitive" trick as $\mathrm{Out}(S_6)$. Conway & Curtis's Miracle Octad Generator (MOG, 1973) realises $M_{24}$ through the Steiner system S(5, 8, 24). Reference: J. Conway et al., ATLAS of Finite Groups (1985).

30.9 PSL₂(𝔽₅) and the icosahedron

The "square-determinant" subgroup of ${\mathrm{PGL}}_2({\mathbb{F}}_5)$ is$${\mathrm{PSL}}_2({\mathbb{F}}_5)={\mathrm{SL}}_2({\mathbb{F}}_5)/\{\pm I\}.$$ Its order is $120/2=60$, and famously${\mathrm{PSL}}_2({\mathbb{F}}_5)\cong A_5$ which in turn is the rotation group $I$ of the icosahedron. Three identifications of the same 60-element group:

The icosahedron has 12 vertices, falling into 6 antipodal pairs; each pair receives a label from ${\mathbb{P}}^1({\mathbb{F}}_5)$. $A_5$ contains elements of order 1, 2, 3, 5: order-5 rotations through opposite vertices (2π/5), order-3 through opposite face centres (2π/3), order-2 through midpoints of opposite edges. Counts: 1 + 15 + 20 + 24 = 60.

This is the subject of Felix Klein's Lectures on the Icosahedron (1884): roots of quintic equations can be expressed via icosahedral functions (a kind of elliptic modular function), trading on the fact that $A_5$ is unsolvable yet has the structure of a 60-element simple group. Klein took the trinity ${\mathrm{PSL}}_2({\mathbb{F}}_5)\cong A_5\cong I$ as the heart of his book.

30.10 Explicit generators and presentation

Two simple Möbius transformations generate ${\mathrm{PSL}}_2({\mathbb{F}}_5)$. Set

with matrices

Check: $T^5=I$ (5 ≡ 0 mod 5), $S^2=-I$ which is trivial in $\mathrm{PSL}$, and $ST:z\mapsto -1/(z+1)$ has order 3 (direct computation gives $(ST{)}^3=\mathrm{id}$). So

This is the von Dyck triangle group $(2,3,5)$, the rotation group of the spherical triangle $\Delta (2,3,5)$ — icosahedral geometry. To upgrade to ${\mathrm{PGL}}_2({\mathbb{F}}_5)\cong S_5$, append one diagonal element

$D$ has determinant 2, not a square in ${\mathbb{F}}_5^{\ast }$, so $D\notin {\mathrm{PSL}}_2$. Then $⟨S,T,D⟩={\mathrm{PGL}}_2({\mathbb{F}}_5)$, doubling 60 → 120. Full presentation: Coxeter–Moser (1965) §6.5.

Cube correspondence: as corner permutations, R is a 4-cycle on 4 corners and U is a staggered 4-cycle, sharing 2 corners. Choose ${\mathbb{P}}^1({\mathbb{F}}_5)$ labels so that

Jaap Scherphuis's labelling gives R = (0 1 2 3) (a 4-cycle on ${\mathbb{P}}^1$ fixing 4 and ∞) and U = (0 4 ∞ 1) (the staggered 4-cycle). The exact labelling is not unique; any choice that turns R, U into a pair of 2×2 mod-5 matrices works.

30.11 Experiment — ⟨R, U⟩ order distribution matches S₅ classes

If $⟨R,U⟩=S_5$, then the order of a random ⟨R, U⟩ word, viewed as a permutation of the 6 corners, should match the conjugacy-class sizes of $S_5$:

Sum: $1+25+20+30+24+20=120$ ✓. Warning: the cycle types above are those of $S_5$ acting on 5 elements; in our 6-point Möbius action, the corresponding cycle types differ. For example a transposition in $S_5$ corresponds to a Möbius transformation acting on 6 points with cycle type (2, 2, 1, 1) — two transpositions and two fixed points. This "cycle type shift" between the two embeddings is the operational core of 30.6's transposition ↔ (2,2,2) swap.

References

• J. Scherphuis, The two-face six-corner puzzle and PGL₂(F₅) — jaapsch.net/puzzles/pgl25.htm. The starting point for this section.
• J. Baez, Some Thoughts on the Number 6 — math.ucr.edu/home/baez/six.html. The classic expository essay on the S₆ outer automorphism, covering synthemes, Mathieu groups, and the icosahedral connection.
• J. Conway, R. Curtis, S. Norton, R. Parker, R. Wilson, ATLAS of Finite Groups (Oxford, 1985). The standard reference for all 26 sporadic simple groups, including maximal-subgroup tables for M₁₁, M₁₂, M₂₄ (in which PGL₂(F₅) appears).
• M. Suzuki, Group Theory I, II (Springer Grundlehren 247, 248, 1982–1986). The standard reference for classical groups (GL, SL, PSL, PGL, PΓL).
• M. Aschbacher, Finite Group Theory (Cambridge Studies in Adv. Math. 10, 2nd ed. 2000). Compact modern survey, background for the classification of finite simple groups.
• S. Lang, Algebra (3rd ed., Springer GTM 211, 2002). General-field GL/PGL/PSL definitions, Chapter 13.
• D. Surowski, Workbook in Higher Algebra. Develops the PGL₂(F₅) action on ℙ¹(F₅) and the cross-ratio as exercises.
• H. S. M. Coxeter & W. O. J. Moser, Generators and Relations for Discrete Groups (Springer Ergebnisse 14, 4th ed. 1980). Standard presentations for triangle groups (2, 3, 5) etc., with icosahedral geometry.
• F. Klein, Vorlesungen über das Ikosaeder (1884, English transl. Lectures on the Icosahedron, Dover 1956). Solving the quintic using PSL₂(F₅) ≅ A₅ ≅ I.
• O. Hölder, "Bildung zusammengesetzter Gruppen," Math. Ann. 46 (1895) — original discovery of $\mathrm{Out}(S_6)$.
• J. J. Sylvester, "Elementary researches in the analysis of combinatorial aggregation," Phil. Mag. 24 (1844). Origin of the duad/syntheme/synthematic-total terminology.