§5

Three invariants (reachability conditions)

Which states are reachable by face turns alone? Exactly these three constraints determine it:

Theorem 5.1 — first law of cubology

A state

(c_{p}, c_{o}, e_{p}, e_{o})

is reachable (i.e. lies in G) if and only if all three hold:

(1)

i = 1 \sum 8 c_{o}^{(i)} \equiv 0 (mod 3)

total corner twist conserved

(2)

i = 1 \sum 12 e_{o}^{(i)} \equiv 0 (mod 2)

total edge flip conserved

(3)

sgn (c_{p}) = sgn (e_{p})

corner-edge parity coupling

Each constraint forbids one physically intuitive move. You cannot twist a single corner (violates 1), flip a single edge (violates 2), or swap two edges without disturbing corners (violates 3). Popping the cube apart and reassembling it sidesteps these — 12 parallel "alternate universes" of unreachable states.

Interactive § Three invariants

Every legal cube state satisfies three constraints. Manually break any one and the state is unreachable — no sequence of face turns can produce it.

alg

twist corner 0flip edge 0swap two edges

Σ co (mod 3)

Σ eo (mod 2)

sgn(cp)

sgn(ep)

✓ Reachable — this state is in G

5.1 Proof: corner orientation sum is conserved

Proof

Claim: applying any generator $X$ preserves $\sum_{i} c_{o}^{(i)} (mod 3)$ .

Verify on the 6 generators:

U, D: these cycle the four U-layer (or D-layer) corners while keeping the U-coloured sticker on the U/D face. So all four $Δ c_{o} = 0$ . ✓
R, L: R cycles four corners. URF → UBR: U sticker rotates "up" → "front-up", $+ 1 (mod 3)$ . UBR → DRB: "up" → "right", $+ 2 (mod 3)$ . By symmetry DRB → DFR = $+ 1$ , DFR → URF = $+ 2$ . Total: $1 + 2 + 1 + 2 = 6 \equiv 0 (mod 3)$ . ✓
F, B: similarly each contributes $Δ (\sum c_{o}) = 6 \equiv 0 (mod 3)$ . ✓

So every generator preserves $\sum_{i} c_{o}^{(i)} (mod 3)$ , and so does any finite product.

∎

5.2 Proof: edge orientation sum is conserved

Proof

Claim: $\sum_{i} e_{o}^{(i)} (mod 2)$ is preserved.

U, D, R, L do not affect any edge's EO (their stickers stay on the same {U/D, L/R} pair). F and B each flip 4 edges, contributing $Δ (\sum e_{o}) = 4 \equiv 0 (mod 2)$ . ✓

Every generator gives $\equiv 0 (mod 2)$ , so $\sum e_{o}$ is a G-invariant.

∎

5.3 Proof: corner-edge parity coupling

Proof

Claim: $sgn (c_{p}) = sgn (e_{p})$ is preserved.

Each face turn cycles 4 corners (a 4-cycle in $c_{p}$ ) and 4 edges (a 4-cycle in $e_{p}$ ). A 4-cycle factors into 3 transpositions, so $sgn = (- 1)^{3} = - 1$ .

Therefore every generator flips $sgn (c_{p})$ and $sgn (e_{p})$ simultaneously. Their ratio $\frac{sgn ( c _{p} )}{sgn ( e _{p} )} = + 1$ stays constant. ✓

∎

Together these three proofs pin G's location inside the free assembly group $S_{8} \times S_{12} \times (Z /3)^{8} \times (Z /2)^{12}$ . The converse — that every state satisfying these three constraints is reachable — is usually established constructively: any working solver is itself a proof of reachability.

Corollary 5.4

The "free assembly space" / G is a 12-element quotient group:

(S_{8} \times S_{12} \times (Z /3)^{8} \times (Z /2)^{12}) / G ≅ Z /3 \times Z /2 \times Z /2

Each cell of the quotient is one "disassembly-only" anomaly: extra CO twist (ℤ/3), extra EO flip (ℤ/2), wrong parity (ℤ/2). That is precisely why a popped-and-rebuilt cube falls into one of 12 "parallel universes", most of which cannot be solved by face turns.

5.4 Conservation laws as cohomology constraints

From a more abstract angle: the three conservation laws are precisely the obstruction classes in the first group cohomology $H^{1} (G, Z / n)$ when viewing the "free assembly space" $F$ as an Abelian extension.

H^{1} (G, (Z /3)^{8}) \supseteq {\sum c_{o} mod 3}, H^{1} (G, (Z /2)^{12}) \supseteq {\sum e_{o} mod 2}

We verified things combinatorially, but the same conclusion drops out of the cohomology-vanishing statement "the six generators all sum to 0 in ℤ/3." This is why "cube invariants," "planar-graph colorings," "topological indices," and "Stiefel–Whitney classes" are siblings in the same abstract family.

5.5 Per-generator verification table

Three tables — for each generator, all three invariant increments vanish modulo the relevant base.

Gen	$Δ (\sum c_{o}) mod 3$	$Δ (\sum e_{o}) mod 2$	$sgn (c_{p}) \cdot sgn (e_{p})$
U	0	0	+1
D	0	0	+1
R	1+2+1+2 = 6 ≡ 0	0	(−1)(−1) = +1
L	6 ≡ 0	0	+1
F	6 ≡ 0	4 ≡ 0	+1
B	6 ≡ 0	4 ≡ 0	+1

Each row directly verifies "this face turn preserves the invariant." Since G is generated by the six face turns, every element does — an automated form of the proofs in §5.1–5.3.

∎

cuberoot.me · Rubik's Cube as a Group · 2026