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Furthermore, hl−1 = γl−1 (gl−1 ) is uniform in Hl−1 and by the previous lemma, this implies gl−1 gl is uniform in Gl−2 . Continuing by induction, we obtain g = g1 g2 . . gl is uniform in G0 = G. In fact, the lemma remains true even if we remove the restriction on the product order of the bij , as we will show in the following theorem. Definition 6 Let Λ = {(i, j) : i = 1, 2, . . l, j = 1, 2, . . ri }. A word w in bij , (i, j) ∈ Λ is complete if bij occurs in w for all (i, j) ∈ Λ. Theorem 7 If B is a Hall basis and w is a complete word, then wα is uniform in G.

Diameter Problem Suppose we have An , Sn and ~~ = An , where S is a set of generators. Conjecture 6. diameter(An , S) < cn2 , c - constant. Also works for Sn . Look at the following weaker versions of this: 1. For the worst case when |S| = 2, we have the following: √ Theorem 7. (Babai, Hetyei) diam < e permuations in Sn . n log n(1+o(1)) . This gives a bound of the maximum order of Lecture 14: 17 October 2001 3 Aim: Find something similar for SL(n, p). Conjecture 8. G - simple ⇒ diam = O((log |G|)c ). ~~

Lemma 5 Let B = (B1 , B2 , . . Bl ), Bi = {bi1 , bi2 , . . biri }, and suppose αij is uniform in {0, 1, . . p − 1}. ri → denotes the product in the order 1, 2 . . l. l Proof: Note that since B is a Hall basis, g can be written as g = g1 g2 . . gl with gi ∈ Gi−1 and hi = γi (gi ) uniform in Hi . Since Hl = Gl−1 , hl = γl (gl ) = gl is uniform in Gl−1 . Furthermore, hl−1 = γl−1 (gl−1 ) is uniform in Hl−1 and by the previous lemma, this implies gl−1 gl is uniform in Gl−2 . Continuing by induction, we obtain g = g1 g2 .