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By Murakami M.

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Proof We have to show that an arbitrary cycle in Ω can be decomposed into quadrangles. Let σ = (x0 , x1 , . . , xn ) be such a cycle of length n, say. Since Ω is bipartite, n is even, say n = 2m. If m = 2, σ is itself a 4-cycle, so we assume that m ≥ 3. One can observe that it is suﬃcient to decompose the cycles such that dΩ (x0 , xm ) = m. The 2-arc π = (xm−1 , xm , xm+1 ) satisﬁes dΩ (x0 , xm−1 ) = dΩ (x0 , xm−1 ) = m − 1. Let Σ be the unique geometric cubic subgraph in Ω containing π. 1 (vi)) there is a unique vertex u in Σ nearest to x0 .

1 Let α : U4 → A[1] , β : U4 → B [1] and ζ : U4 → Z [1] be the [1] isomorphism commuting with the action of L . Then the following key commutator relation holds [α(w1 ), β(w2 )] = ζ(w1 ∧ w2 ) for all w1 , w2 ∈ U4 . 2) Q[1] = Z [1] A[1] B [1] , Q[0] = Z [1] A[1] , and H [01] = Q[1] L[1] . Finally, R is of order 2 generated by the element t0 which is the orthogonal transvection with respect to the unique non-singular vector u + v of W2 . Then t0 centralizes both Z [1] and L[1] , and conjugates s u,w onto s v,w for every w ∈ U4# .

Proof Let S be a Sylow 2-subgroup in L. Then |S| = 8, S intersects trivially the stabilizer of a point from P8 and hence S acts transitively on P8 . Since S must stabilize a non-zero vector in each L-submodule, we conclude that P8c is the unique minimal L-submodule and dually P8e is the only maximal submodule which gives (i). 6)). Hence in order to establish (ii) it is suﬃcient to show that I1 : L contains at most two classes of complements. Let L(1) be a complement which is not in the class of L.