6 1. TH E CALCULUS O F SWEEDLE R POWER S LEMMA. For natural numbers m i , . . . , rrik and m j , . . . , raj, we have ( Pmi,...,m f e ,m / 1 ,...,mJ = Pn.n' ° \(Pmi,...,rrik * i Pm'1,...,m,l) where n := mi 7712 ... ra^ and n' := ra^ ra^ ... raj. Of course, this can also be seen from the explicit formulas, as we have = " ' ( V m 1 , . . . , m f c ( « l , - - - , i f c ) - 1 ) + Pm'1,...,m'l(i'l, ,i'l) k I = £ ( t r - l)n'n r +i + £)(** - X K+i + * r—1 s = l ~ ( /^mi,...,7nfc,?n/1,...,mJ V^l? 5 ^kt H ' 5 2jJ where we have set n^+i = nj + 1 = 1. 1.2. We now consider the union oo S:=[jS n 7 1 = 1 of all symmetric groups. Note that the different symmetric groups are disjoint, which implies that for an element a G S we can say exactly to which symmetric group Sn it belongs. This unique n will be called the degree of a. On the set 5, we introduce a product as follows: For a G Sn and r G ^m? we define cr r G 5 m n to be the unique permutation that makes the following diagram commutative: Wj).r(i) ) Y Y Explicitly, this permutation is given by the formula a r((i - \)n + j) = (a(j) - l)ra + r(z) This product turns 5 into a monoid: PROPOSITION. S is a monoid with unit element id^ G S\. PROOF . Suppose that n, ra, and p are natural numbers. For p G Sp, a G 5 n , and r G 5 m , we have to show that (p-cr) r = p- (a r). For this, note that the following
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