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May 1, 2020 at 7:28 comment added A. Gupta 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).
May 1, 2020 at 7:12 comment added A. Gupta 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 $-\{e_3 \wedge e_4\}$.
Apr 30, 2020 at 17:22 comment added Dan Petersen The only exception is when $n=4$ in which case a product of two disjoint transposition also works.
Apr 30, 2020 at 9:51 vote accept A. Gupta
Apr 30, 2020 at 9:45 history edited A. Gupta CC BY-SA 4.0
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Apr 30, 2020 at 7:23 answer added Lior Silberman timeline score: 2
Apr 30, 2020 at 6:16 comment added A. Gupta Yes, we mean exactly the restriction to $\bar B$ ("latter action" is now made explicit in the question).
Apr 30, 2020 at 6:13 history edited A. Gupta CC BY-SA 4.0
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Apr 30, 2020 at 5:51 comment added A. Gupta Please, note a modification in the question.
Apr 30, 2020 at 5:49 history edited A. Gupta CC BY-SA 4.0
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Apr 30, 2020 at 5:17 history edited A. Gupta CC BY-SA 4.0
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Apr 30, 2020 at 3:59 history asked A. Gupta CC BY-SA 4.0