By Pak I.

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Extra info for Combinatorics, Probability and Computations on Groups

Sample text

With a 1 in the ith diagonal.  ..      .. 0       0 ··· 0 1 Thus |Bi | = n − i. Now we have to check B is a Hall basis for U(n, p). This is “obvious” since we know that <γi (Bi )> = Hi since, firstly = Gi , where Gi consists of 0’s everywhere below the ith diagonal except the main diagonal, and Hi is the quotient Gi−1 /Gi , so Hi ∼ = (Zp )n−i . For the mixing time, we  1 0 0 1  Eij (α) =  . ..  .. 0 ··· know Xt = Ei1 j1 (α1 ) · Ei2 j2 (α2 ) · . . · Eit jt (αt ) by definition since  ··· 0 α 0  ..

Theorem 6. If G is a finite p-group, then G is nilpotent. Example 2. The group of upper triangular matrices U (n, p) of Fp is a p-group, of the lower central series yields:    1 0 ∗ ∗ ··· ∗ 1 0 0 ∗  1 0 0  1 0 ∗ · · · ∗      1 0 · · · ∗ 1 0    G1 =  , G =   . 2 . .. . ..    .       1 0 1 hence nilpotent. Calculation ··· ··· ··· .. 1  ∗ ∗  ∗  ..   0 1 and so forth, so that Gn−1 consists only of the identity matrix. Interestingly enough, we can deduce from this that Hi Zn−1 .

The following lemma shows that for a Hall basis B, the distribution of B α over α is uniform in H. Lemma 2 If B = {b1 , b2 , . . br } is a Hall basis in H, then αr 1 α2 P rα (bα 1 b2 . . br = h) = 1 ∀h ∈ H |H| Proof: Consider H as a vector space. Since B is a Hall basis, it is a spanning set and contains a basis, say b1 , b2 , . . bk . Then for uniform αi ∈ {0, 1, . . p − 1} , we have α α αk k+1 k+2 αr 1 α2 bα 1 b2 . . bk bk+1 bk+2 . . br , uniform in H which is uniform in H. Recall that G is nilpotent if some Gl in the lower central series G = G0 ⊃ G1 ⊃ G2 ⊃ G3 ⊃ · · · ⊃ Gl = {1} is the identity element (where Gi = [Gi−1 , G] for i = 1, 2, 3 .