== Theorem 1.2(a) over F_9=F_3[t]/(t^2+1) ; iota = (0, 1) ; iota^2 = -1 checked
|SL_2(9)| by enumeration from t12(1),t21(1),t12(i),t21(i): 720
|H| = |<t12(1), t21(i)>| = 120
element orders (order: count): {1: 1, 2: 1, 3: 20, 4: 30, 5: 24, 6: 20, 10: 24}
matches the note's list / SL_2(5): True
H cap U12 = t12(['(0, 0)', '(1, 0)', '(2, 0)']) == t12(F_3): True
H cap U21 = t21(['(0, 0)', '(0, 1)', '(0, 2)']) == t21(F_3 i): True
orbit-Schreier method: |H| = 120  closed12: True  closed21: True  ABA<=A: False  BAB<=B: False
centre of H: ['[[(2, 0),(0, 0)],[(0, 0),(2, 0)]]', '[[(1, 0),(0, 0)],[(0, 0),(1, 0)]]'] ; involutions: ['[[(2, 0),(0, 0)],[(0, 0),(2, 0)]]']
|[H,H]| = 120 (perfect: True )
=> H is a perfect group of order 120 with centre {+-I} of order 2; H/Z is perfect of order 60, hence A_5;
   a perfect central extension of A_5 by C_2 is SL_2(5) (Schur multiplier of A_5 is C_2). So H ~ SL_2(5).
elements of H of the form [[0,t],[-1/t,0]] (Levchuk 1983 says H contains one): 4 e.g. [[(0, 0),(1, 1)],[(1, 2),(0, 0)]]
ab = [[(1, 1),(1, 0)],[(0, 1),(1, 0)]]  order 10 ; trace = (2, 1)

== Remark 4.1: A0 = F_3 and the four lines B0 of F_9
  B0 = F_3*(0, 1) |H| =  120 ; closed12=True closed21=True ; ABA<=A: False ; BAB<=B: False (orbit-method |H| = 120)
  B0 = F_3*(1, 0) |H| =   24 ; closed12=True closed21=True ; ABA<=A: True ; BAB<=B: True (orbit-method |H| = 24)
  B0 = F_3*(1, 1) |H| =  720 ; closed12=False closed21=False ; ABA<=A: False ; BAB<=B: False (orbit-method |H| = 720)
  B0 = F_3*(1, 2) |H| =  720 ; closed12=False closed21=False ; ABA<=A: False ; BAB<=B: False (orbit-method |H| = 720)

== Theorem 1.2(b) over F_4=F_2[t]/(t^2+t+1) ; omega^2 = omega+1 checked
|<a,b>| = 10 ; a^2 = 1: True ; b^2 = 1: True
g = ab = [[(1, 1),(1, 0)],[(0, 1),(1, 0)]] ; expected [[1+w,1],[w,1]]: True
trace(g) = (0, 1)  det(g) = (1, 0)  order(g) = 5
element orders: {1: 1, 2: 5, 5: 4} (D_10: {1:1, 2:5, 5:4})
a g a^{-1} = g^{-1} (dihedral relation): True
(X^2+wX+1)(X^2+w^2X+1) coefficients (low->high): ['(1, 0)', '(1, 0)', '(1, 0)', '(1, 0)', '(1, 0)'] == X^4+X^3+X^2+X+1: True
H cap U12 = t12(['(0, 0)', '(1, 0)']) ; H cap U21 = t21(['(0, 0)', '(0, 1)'])
  == t12(F_2): True  == t21(F_2 w): True
  omega in F_2 ? False  (so A0 B0 A0 not in A0)

== Remark 4.1 over F_4: A0 = F_2, the three lines B0
  B0 = F_2*(0, 1) |H| =  10 orders {1: 1, 2: 5, 5: 4} ; closed12=True closed21=True ; ABA<=A: False
  B0 = F_2*(1, 0) |H| =   6 orders {1: 1, 2: 3, 3: 2} ; closed12=True closed21=True ; ABA<=A: True
  B0 = F_2*(1, 1) |H| =  10 orders {1: 1, 2: 5, 5: 4} ; closed12=True closed21=True ; ABA<=A: False

== Cross-check: orbit/Schreier closedness test vs full enumeration, all pairs of nonzero subgroups
  F_4=F_2[t]/(t^2+t+1)   pairs= 16  disagreements=0
  F_9=F_3[t]/(t^2+1)     pairs= 25  disagreements=0
  F_25=F_5[t]/(t^2+3)    pairs= 49  disagreements=0

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