Past exam of the mathematics course of the University of Oxford / 2014 / Prelims / Mathematics I / 6 / b

Conjugation of a permutation relabels each symbol by the inverse conjugating permutation. Specifically,

τ^{- 1} (a_{1} a_{2} \dots a_{n}) τ = (τ^{- 1} (a_{1}) τ^{- 1} (a_{2}) \dots τ^{- 1} (a_{n})) .

(1)

To verify the direction of the inverse, apply the left side to

τ^{- 1} (a_{j})

τ

first takes it to

a_{j}

, the original cycle takes it to

a_{j + 1}

, and

τ^{- 1}

takes it to

τ^{- 1} (a_{j + 1})

, with cyclic indexing. Any symbols fixed by the original permutation are similarly relabelled and remain fixed. More generally, conjugate every cycle of a disjoint-cycle decomposition in this way; disjointness is preserved because

τ^{- 1}

is a bijection.

ii

 0  0

Solution

 0  0

By the preceding relabelling formula, every conjugate of

(123)

is a three-cycle, and any three-cycle on four symbols can be obtained by a suitable relabelling. Therefore

C ((123)) = {(123), (132), (124), (142), (134), (143), (234), (243)} .

(1)

There are eight elements, agreeing with the count

(3 4) \cdot 2

Likewise, conjugation preserves the pattern of two disjoint transpositions, and every partition of four symbols into two pairs is realized by relabelling. Thus