Much as I like Burnside's Lemma, induced (permutation) representations, and other parts of group theory, I can't resist pointing out an alternative argument that uses essentially no group theory but relies on the fact that expectation (of random variables) is linear. Since is the number of fixed-points of , its square is the number of fixed ordered pairs , where of course fixing a pair means fixing both its components. So the in the question is the average number of fixed pairs of a permutation , in other words the expectation (with respect to the uniform probability measure on ) of the random variable "number of fixed pairs." That random variable is the sum, over all pairs , of the indicator variable whose value at any permutation is 1 or 0 according to whether fixes and or not. So is the sum, over all , of the expectations of these , and these expectations are just the probabilities that a random permutation fixes and . For each of the pairs where , that probability is , so all these together contribute 1 to the sum. For each of the remaining pairs, the probability is (namely, probability to fix and conditional probability to fix given that is fixed). So these pairs also contribute 1 to the sum, for a total of 2.
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