action of symmetric group on the second exterior power

Question

Let $e_i \wedge e_j \ (i < j)$ be a basis for the $\mathbb Z$-module $\wedge^2 \Gamma$, where $\Gamma = \mathbb Z^n$. Clearly $S_n$ acts on the module $\wedge^2 \Gamma$ via $$\pi(e_i \wedge e_j) = e_{\pi(i)} \wedge e_{\pi(j)} \ \ \ \forall \pi \in S_n.$$ By restriction this induces an action on the subset $\bar B = \{ \epsilon e_i \wedge e_j \ (i < j), \ \epsilon \in \{-1, 1\} \}$.

Which (non-trivial) cyclic subgroups of $S_n$ have maximal number of orbits in this action on $\bar B$. The answer seems to be the subgroups generated by transpositions $\pi = (ij)$. But can there be other permutations $\pi$ that are not transpositions but with the same number of orbits?

Answer 1

By the Lemma that is not Burnside's, the number of orbits is the average number of fixed points. An element fixes $\epsilon e_i \wedge e_j$ iff it fixes both $i,j$ (because if it swaps them it reverses the sign, and otherwise it won't even preserve the span $\mathbb{Z} e_i \wedge e_j$. It follows that $\sigma \in S_n$ will have $2\binom{\#\mathrm{Fix}(\sigma)}{2}$ fixed points on $\bar{B}$ because we need to choose pairs $(i,j)$ in its fixed point set.

Now a transposition has the largest number of fixed points of any non-identity element. Accordingly let $G < S_n$ be any non-trivial subgroup and let $C<S_n$ be the subgroup generated by a transposition. Then we have

$ \# \bar{B}/G = \frac{1}{\# G} \sum_{\sigma\in G} 2\binom{\#\mathrm{Fix}(\sigma)}{2} \leq \frac{1}{\# G} n(n-1) + \left(1-\frac{1}{\# G}\right)(n-2)(n-3) = (n-2)(n-3) + \frac{1}{\# G} (4n-6) \leq (n-2)(n-3) + \frac{1}{\# C} (4n-6) = \# \bar{B}/C \,.$

It follows that the number of orbits of $C$ is maximal, with equality iff $G$ is conjugate to $C$.