Product-one counts in the symmetric group on three letters #
In S₃ the three-cycles form a conjugacy class of size 2 and the transpositions one of size 3.
The two counts below are decided by kernel computation and factor as
TauCeti.card_productOneTriples predicts: 6 = 3 · 2 at a three-cycle and two transpositions, and
0 = 3 · 0 at three transpositions, the second factor being the structure constant. On the
character side the class sizes contribute 2 · 3 · 3 / 6 = 3 and 3 · 3 · 3 / 6, so the character
sums are 2 and 0.
Main results #
TauCeti.card_productOneTriples_threeCycle_transpositionandTauCeti.card_productOneTriples_transposition: the product-one counts.TauCeti.sum_characterTable_threeCycle_transpositionandTauCeti.sum_characterTable_transposition: the corresponding character sums.
A transposition of Fin 3 factors as a transposition times a three-cycle in two ways.
Six product-one triples in S₃ at a three-cycle and two transpositions. A triple
(x, y, z) with x a three-cycle and y, z transpositions satisfies z * y * x = 1 exactly
when x = y * z, which is a three-cycle precisely when y ≠ z; there are six such ordered
pairs.
No product-one triple in S₃ has all three entries transpositions. A product of two
transpositions of Fin 3 is the identity or a three-cycle, never a transposition.
The character side of the Frobenius formula in S₃, at a three-cycle and two
transpositions: the sum over the irreducible characters is 2, the count of six product-one
triples divided by the class-size factor 2 · 3 · 3 / 6 = 3.
The character side of the Frobenius formula in S₃, at three transpositions: the sum over
the irreducible characters is 0, since there is no product-one triple of transpositions and the
class-size factor 3 · 3 · 3 / 6 is nonzero.