Show that if p is prime, then Sp = { a, b | a,b∈Sp, a is a transposition, b is a p-cycle}.
Proof: Let a1 and a2 be the indices transposed by a. Let c1 and c2 be the indices that b permutes a1 and a2 to, respectively. Therefore, bab-1 = ( c1 c2 ) = c. As we can see, bcb-1 = ( d1 d2 ), where d1 and d2 are the indices that b permutes c1 and c2 to. In this fashion, we can construct all of the transpositions that are of the same distance away in b as a1 is to a2. Call this distance m.
By using induction, all of the linked indices above and alike are generated transpositions. Every transposition of distance m is within { a, b | a,b∈Sp, a is a transposition, b is a p-cycle}.
Take the two generated transpositions ( 1 x ) and ( 1 y ), where x is going one direction across b and y another. If x and y were the same element, that would imply that m ≡ -m mod p, so p | 2m, a contradiction since ( p,m ) = 1.
( 1 y ) ( 1 x ) ( 1 y ) = ( x y ), a transposition of distance 2m. Using the same method as demonstrated above, every transposition of distance 2m is within { a, b | a,b∈Sp, a is a transposition, b is a p-cycle}. Take ( z y ) as a transposition of distance m, where ( x z ) would be of distance 3m rather than m.
( z y ) ( x y ) ( z y ) = ( x z ), a transposition of distance 3m. Continuing this process, every transposition of distance jm is within { a, b | a,b∈Sp, a is a transposition, b is a p-cycle} for j∈Z+. Since ( p,m ) = 1, there exists integers k and i such that ip + km = 1, so for all integers g, (kg)m ≡ g mod p. So there exists f such that fm ≡ g mod p, for any integer g. Since two sets of transpositions of distances that are congruent to each other modulo p are in fact the same set, this implies that every transposition of any distance is contained in { a, b | a,b∈Sp, a is a transposition, b is a p-cycle}. Since every permutation in any symmetric group is rewritable as a product of transpositions, this implies that Sp = { a, b | a,b∈Sp, a is a transposition, b is a p-cycle}.
∎
Looking closely, one can see that the primality of p was not specially invoked in this proof. The only assumption is that the distance of the transposition a is relatively prime to p, which may be composite.
( 1 y ) ( 1 x ) ( 1 y ) = ( x y ), a transposition of distance 2m. Using the same method as demonstrated above, every transposition of distance 2m is within { a, b | a,b∈Sp, a is a transposition, b is a p-cycle}. Take ( z y ) as a transposition of distance m, where ( x z ) would be of distance 3m rather than m.
( z y ) ( x y ) ( z y ) = ( x z ), a transposition of distance 3m. Continuing this process, every transposition of distance jm is within { a, b | a,b∈Sp, a is a transposition, b is a p-cycle} for j∈Z+. Since ( p,m ) = 1, there exists integers k and i such that ip + km = 1, so for all integers g, (kg)m ≡ g mod p. So there exists f such that fm ≡ g mod p, for any integer g. Since two sets of transpositions of distances that are congruent to each other modulo p are in fact the same set, this implies that every transposition of any distance is contained in { a, b | a,b∈Sp, a is a transposition, b is a p-cycle}. Since every permutation in any symmetric group is rewritable as a product of transpositions, this implies that Sp = { a, b | a,b∈Sp, a is a transposition, b is a p-cycle}.
∎
Looking closely, one can see that the primality of p was not specially invoked in this proof. The only assumption is that the distance of the transposition a is relatively prime to p, which may be composite.
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