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Oct 2, 2012 at 0:38 comment added Richard Stanley For the generalization to commuting $k$-tuples of elements in $S_n$ and related results, see Exercise 5.13 in Enumerative Combinatorics, vol. 2.
Sep 29, 2012 at 2:28 history edited Geoff Robinson CC BY-SA 3.0
Noted answer to more general version of the question
Sep 23, 2012 at 18:10 comment added Mark Wildon For symmetric groups there is the beautiful identity $$ \prod_{j=1}^\infty (1-u^j)^{-\sigma(j)} = \sum_{n=1}^\infty \frac{T(n)}{n!} u^n $$ where $T(n)$ is the number of triples of commuting elements in the symmetric group $S_n$ and $\sigma$ is the sum of divisors function. See arxiv.org/abs/1203.5079 by John Britnell for an elementary proof.
Sep 23, 2012 at 16:15 history edited Geoff Robinson CC BY-SA 3.0
Noted equivalent formulation
Sep 23, 2012 at 15:48 history answered Geoff Robinson CC BY-SA 3.0