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?

  • $\begingroup$ Please, note a modification in the question. $\endgroup$
    – A. Gupta
    Apr 30 '20 at 5:51
  • $\begingroup$ Yes, we mean exactly the restriction to $\bar B$ ("latter action" is now made explicit in the question). $\endgroup$
    – A. Gupta
    Apr 30 '20 at 6:16
  • 1
    $\begingroup$ The only exception is when $n=4$ in which case a product of two disjoint transposition also works. $\endgroup$ Apr 30 '20 at 17:22
  • $\begingroup$ Please note: For $n = 4$ there are $7$ orbits for the group $\langle (12) \rangle$, namely, $\{\pm e_1\wedge e_2\}$, $\{e_1 \wedge e_3, e_2 \wedge e_3 \}$, $\{-e_1 \wedge e_3, -e_2 \wedge e_3 \}$, $\{e_1 \wedge e_4, e_2 \wedge e_4 \}$, $\{-e_1 \wedge e_4, -e_2 \wedge e_4 \}$, $\{e_3 \wedge e_4 \}$ and $-\{e_3 \wedge e_4\}$. $\endgroup$
    – A. Gupta
    May 1 '20 at 7:12
  • $\begingroup$ For the action of $\langle (12)(34) \rangle$ there are six orbits only: $\{e_1 \wedge e_2, -e_1 \wedge e_2 \}$, $\{e_3 \wedge e_4, -e_3 \wedge e_4\}$, $\{e_1 \wedge e_3, e_2 \wedge e_4 \}$, $\{-e_1 \wedge e_3, -e_2 \wedge e_4 \}$, $\{e_1 \wedge e_4, e_2 \wedge e_3 \}$ and $\{-e_1 \wedge e_4, -e_2 \wedge e_3 \}$. The reason behind this is that $\langle (12) \rangle$ fixes more points in $\bar B$ (on average). $\endgroup$
    – A. Gupta
    May 1 '20 at 7:28

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 $C$.


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