PASS an antidiagonal linear map preserving O exists (found by search)  [J = [[0, 0, 0, 1], [0, 0, 1, 0], [0, 1, 0, 0], [1, 0, 0, 0]]]
PASS (Z1) |S| = q^2(q-1)(q^2+1) = 29120  [|S| = 29120]
PASS S is 2-transitive on the 65 points
PASS normal closure of an element of each order is S  [element orders [1, 2, 4, 5, 7, 13]]
PASS S is simple (every nontrivial class has normal closure S)  [11 classes, sizes [1, 455, 1820, 1820, 2240, 2240, 2240, 4160, 4160, 4160, 5824]]
PASS (Z1) q^2+1 = (q+r0+1)(q-r0+1)
PASS phi preserves O, has order 3, normalizes S, and is not in S
PASS (Z2) U = {U(a,b)} is a subgroup of order q^2 = 64, a Sylow 2-subgroup, nonabelian
PASS (Z2) N_S(U) = B = U:H, |B| = 448, H cyclic of order 7
PASS (Z2) H acts fixed-point-freely on U
PASS (Z2) phi normalizes U and H and acts on H as h_k -> h_(k^2)
PASS (Z3) t=5: T = C_S(x) cyclic of order 5, C_S(y) = T for all 1 != y in T, |N_S(T)| = 20, N_S(T)/T cyclic of order 4
PASS (Z3) t=7: T = C_S(x) cyclic of order 7, C_S(y) = T for all 1 != y in T, |N_S(T)| = 14, N_S(T) dihedral
PASS (Z3) t=13: T = C_S(x) cyclic of order 13, C_S(y) = T for all 1 != y in T, |N_S(T)| = 52, N_S(T)/T cyclic of order 4
PASS (Z3) 5, 7, 13 are the odd primes of |S| and divide exactly one of q-1=7, q-r0+1=5, q+r0+1=13, pairwise coprime
PASS (Cl') N_S(U) = B soluble
PASS Lemma 4.1: C_S(phi) has order 20 = |Sz(2)| (elements of orders 1,2,4,5: the Frobenius group 5:4)
PASS Lemma 4.1: C_S(phi^2) = C_S(phi)
PASS Lemma 4.1: the elements of order 3 of S phi^1 number |S|/|C_S(phi)| = 1456 (one S-class)  [1456 elements of order 3]
PASS Lemma 4.1: the elements of order 3 of S phi^2 number |S|/|C_S(phi)| = 1456 (one S-class)  [1456 elements of order 3]
PASS Lemma 4.2: C_B(U) = Z(U) of order 8, F(B) = U
PASS Lemma 4.2: coset S phi^1, p=3: X = U is a (b2)-witness (O_p(B)=1; all 448 elements normalizing U act nontrivially on B/F*(B))
PASS Lemma 4.2: coset S phi^1, p=5: X = U is a (b2)-witness (O_p(B)=1; all 448 elements normalizing U act nontrivially on B/F*(B))
PASS Lemma 4.2: coset S phi^1, p=7: X = U is a (b2)-witness (O_p(B)=1; all 448 elements normalizing U act nontrivially on B/F*(B))
PASS Lemma 4.2: coset S phi^1, p=11: X = U is a (b2)-witness (O_p(B)=1; all 448 elements normalizing U act nontrivially on B/F*(B))
PASS Lemma 4.2: coset S phi^1, p=13: X = U is a (b2)-witness (O_p(B)=1; all 448 elements normalizing U act nontrivially on B/F*(B))
PASS Lemma 4.2: coset S phi^2, p=3: X = U is a (b2)-witness (O_p(B)=1; all 448 elements normalizing U act nontrivially on B/F*(B))
PASS Lemma 4.2: coset S phi^2, p=5: X = U is a (b2)-witness (O_p(B)=1; all 448 elements normalizing U act nontrivially on B/F*(B))
PASS Lemma 4.2: coset S phi^2, p=7: X = U is a (b2)-witness (O_p(B)=1; all 448 elements normalizing U act nontrivially on B/F*(B))
PASS Lemma 4.2: coset S phi^2, p=11: X = U is a (b2)-witness (O_p(B)=1; all 448 elements normalizing U act nontrivially on B/F*(B))
PASS Lemma 4.2: coset S phi^2, p=13: X = U is a (b2)-witness (O_p(B)=1; all 448 elements normalizing U act nontrivially on B/F*(B))
PASS Lemma 4.3: coset S phi^1, p=2, r=3 (3 does not divide q-1=7): t=7 is a ppd of 2^3-1, C_S(Q_7) = T of order 7, no element of the coset centralizes Q_7 (b1)
PASS Lemma 4.3: coset S phi^2, p=2, r=3 (3 does not divide q-1=7): t=7 is a ppd of 2^3-1, C_S(Q_7) = T of order 7, no element of the coset centralizes Q_7 (b1)
PASS (Cl') N_S(Q_7) <= N_S(T) soluble; N_S(Q_5), N_S(Q_13) soluble
time 24.0 s
SUMMARY: 34 checks, 0 failed
ALL CHECKS PASSED
