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I am familiar with the formula for computing the size of a conjugacy class of $S_n$ given its cycle type, but other than guessing and checking I do not know how to find either:

  1. The size of the largest conjugacy class
  2. The cycle type of the largest conjugacy class.

Is there any simple way of computing these?

Diffycue
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1 Answers1

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For each cycle type, there is a formula of how many elements are in this conjugacy class, e.g., for $(123\cdots n)$ we see that its conjugacy class has size $(n-1)!$. However, the conjugacy class of $(123\cdots n-1)(n)$ is bigger for $n\ge 3$, namely $$ (n-1)!+(n-2)!=\frac{n!}{n-1}. $$

This is the largest size, if you look at the Conjugacy class size formula in symmetric group. I don't know a simple way of computing it, since we have $p(n)$ conjugacy classes, i.e., cycle types, which grows rapidly with $n$. As for the computation of the size of the largest conjugacy class, see OEIS. It is mentioned, too, that this is the "the maximum entry in each row of A036039".

Dietrich Burde
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  • The OEIS entry A001048 (to which you linked) claims that the conjugacy class of $(123 \cdots n-1)(n)$ is maximal, but gives no proof. However, maximizing the size of the conjugacy class is equivalent to minimizing the denominator $\prod_j j^{a_j} (a_j)!$ in that formula, where $a_1 + 2a_2 + \cdots + na_n = n$. You're looking at $a_1 = 1, a_{n-1} = 1$, and $a_j = 0$ for all other $j$, giving the product $n-1$. That this is the largest seems to be folklore, e.g. Stanley's comment at https://mathoverflow.net/questions/84271/distribution-of-the-sizes-of-conjugacy-classes-in-the-symmetric-group – Michael Lugo Oct 16 '17 at 15:07
  • @MichaelLugo Thank you for your helpful comment, and the MO-link with Stanley's comment. I was trying to find some written proof. – Dietrich Burde Oct 16 '17 at 15:13