By Pak I.

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K = k log k + O(k) k−1 k−2 We’ll fill in more details in the next lecture. 317 Combinatorics, Probability, and Computations on Groups 1 October 2001 Lecture 10 Lecturer: Igor Pak Scribe: Igor Pavlovsky Coupon Collector’s Problem Recall that the problem concerns a prudent shopper who tries, in several attempts, to collect a complete set of N different coupons. Each attempt provides the collector with a coupon randomly chosen from N known kinds, and there is an unlimited supply of coupons of each kind.

By symmetry, we reach m 4 and 4 each with probability 1/2, regardless of the time it takes us to get there. From that point, now we have to go a distance m 8 either forward or backward, and again both happen with equal probability, regardless of the time it takes to get there. So now it is equally likely that we stand at any of those four points. When we’ve finished all but the last stage of the protocol, we have an equal chance of being at any odd-numbered point, regardless of the time it has taken us to get there.

From here, proving that G/[G, G] is abelian is quite straightforward: let a, x ∈ G, and let [a] and [x] be the associated elements of G/[G, G]. Then [x][a] = [xa] = [xa(a−1 x−1 ax)] = [ax] = [a][x] Finally, we shall discuss application of these properties to nilpotent groups. Definition 4. The lower central series of a group G is the chain of groups G0 G1 G2 · · · defined by G0 = G and Gi+1 = [G, Gi ]. Group G is called nilpotent if some G in the lower central series of G is trivial. Let Hi = Gi−1 /Gi .