If sigma is an element of S_n, then sigma has a cycle decomposition into disjoint cycles of various lengths (let us include 1-cycles). Since disjoint cycles commute, the shape of the element is determined by the lengths of the various cycles, which we can assume are put in decreasing order. Any two elements with the same cycle shape are conjugate, so the conjugacy classes are determined by writing n (=52 say) as a sum of decreasing integers.

1. Find the conjugacy class in S_52 with the largest number of elements.

Here is a worked solution (by exhaustion) for S_3. The partitions of 3 are given by 3 = 3, 3 = 2 + 1, and 3 = 1 + 1 + 1. The first partition corresponds to the 3-cycles in S_3, which are

(1,2,3), (1,3,2)

Note that this list consists ofall of the 3-cycles (e.g. (3,2,1) = (1,3,2)). For 3 = 2 + 1, we have

(1,2)(3), (1,3)(2), (2,3)(3).

Finally, for 3 = 1 + 1 + 1 we have:

(1)(2)(3)

which is the identity. So the size of the conjugacy classes is 2 for 3=3, 3 for 3=2+1, and one for 3=1+1+1, so the answer in this case would be that the largest conjugacy class has size 3.

Note that there are 52! = 80658175170943878571660636856403766975289505440883277824000000000000 elements of S_52 and 281589 conjugacy classes, so there must be at least one conjugacy class of size

52!/281589 ~ 1.5181... x 10^62.

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