# Smith normal form of conjugacy class actions

This question was inspired by Smith Normal Form of a Cayley Graph of the Symmetric Group. Let $$\mathbb{Q}S_n$$ denote the group algebra over $$\mathbb{Q}$$ of the symmetric group $$S_n$$. Identify a conjugacy class $$K_\lambda$$ (where $$\lambda$$ is a partition of $$n$$) with the sum of its elements in $$\mathbb{Q}S_n$$. Let $$Z_n$$ denote the center of $$\mathbb{Q}S_n$$, so a $$\mathbb{Q}$$-basis for $$Z_n$$ consists of the sums $$K_\lambda$$. $$K_\lambda$$ acts on $$Z_n$$ by left multiplication. What is the Smith normal form (SNF) over $$\mathbb{Z}$$ of the matrix of this action with respect to the basis $$\{K_\mu\}$$?

Of special interest is the case $$\lambda=(n)$$, so $$K_{(n)}$$ is the class (or sum) of $$n$$-cycles. There are $$n$$ nonzero diagonal elements of the SNF. (This is easy to see.) I have computed them up to $$n=13$$. The $$k$$th diagonal entry ($$0\leq k\leq n-1$$) is $$k!$$ times a rational number with small numerator and denominator. The largest diagonal entry is always $$(n-1)!$$. For instance, when $$n=9$$ the SNF is $$0!,\ 2\cdot 1!,\ 2!,\ \frac 23 3!,\ 4!,\ 2\cdot 5!,\ \frac 136!,\ 2\cdot 7!,\ 8!.$$ When $$n=12$$ the SNF is $$0!,\ 1!,\ 2!,\ \frac 133!,\ \frac 124!,\ 5!,\ 2\cdot 6!,\ 7!,\ \frac 128!,\ \frac 139!,\ 10!,\ 11!.$$

Two specific conjectures:

(a) If $$n$$ is an odd prime, then the nonzero SNF terms are $$k!$$ for $$k$$ even and $$2\cdot k!$$ for $$k$$ odd, where as above $$0\leq k\leq n-1$$.

(b) If $$n$$ is twice an odd prime, then the nonzero SNF terms are $$k!$$ for all $$0\leq k\leq n-1$$, except that $$(n/2)!$$ is omitted, and $$(\frac n2-1)!$$ appears twice.

Possibly with more data one could make a conjecture for all $$n$$.

Note that my question makes sense for any finite group. Can anything be said about the SNF in this generality?