 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2014/prelims/mathematics-i/6/b/ii/solution

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)} .

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

C ((12) (34)) = {(12) (34), (13) (24), (14) (23)} .

These have the respective cycle types

(3, 1)

and

(2, 2)

S_{4}

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