FINDING N-TH ROOTS IN NILPOTENT GROUPS AND ...

584 S. Sze, D. Kahrobaei, R. Dambreville, M. Dupas
unless all occurrences of aj+1 (or a0) in X have been used in the ordering,
in which case we use the next am not already exhausted.
Now we partition the multiset X into submultisets S0,S1,... Sq−1 with k elements
and submultiset Sq with r elements by the following rule:
• For i = 0,1,... ,q − 1, Si = {bik,bik+1,... ,b ik+(k−1)}
• Sq = {aqk,aqk+1,... ,a qk+(r−1)}
Note that Sq may be empty (when r = 0) and S0 contains a0, a1, ..., ak−1.
In addition, the set of distinct elements of Si is a subset of the set of distinct
elements of Si−1.
Next, consider the multiplication table for |X| as a union of tables SiSj
for 0 ≤ i,j ≤ q. In the table SiSj, the number of occurrences of 1 is at least
min{|Si|, |Sj|}, since if i ≤ j, each element of Sj is in Si, and the square of any
element of G/M is equal to 1 (lemma 5.9). Therefore, each row of SiSj contains
an occurrence of 1. Also,
min{|Si|, |Sj|} =
k if i,j < k
r otherwise
Hence, the occurrences of 1 in the multiplication table for |X|, which equals
|M|, is bounded by
|M| =
We have
q
i=0 j=0
q
min{|Si|, |Sj|} = kq 2 + rq + r(q + 1) = kq 2 + 2rq + r
|G| = |X| 2 = (kq + r) 2 = k 2 q 2 + 2kq + r 2
[G : M] ≤ k2 q 2 + 2kq + r 2
kq 2 + 2rq + r = k k2 q 2 + 2kq + r 2
k 2 q 2 + 2krq + kr
because r < k. Now, since the coset M does not contain elements of X, we
also know that [G : M] > k (since there are k cosets of M in X), which is a
contradiction.
≤ k