I have a finite alphabet $\{e_1, e_2, \cdot \cdot\cdot, e_N, a_1, a_2, \cdot\cdot\cdot, a_n \}$ where we pair $e_i$ and $a_i$ as ``opposites'' - like opposite vertexes on a regular $2N$ sided polygon but we are not restricted to the dihedral group. There is some transitive finite group of allowed transformations on these elements, but it could be any group as long as its transitive on the set. My question is: can I always realise the transformation that swaps every $e_i \leftrightarrow a_i$, i.e. swaps every element with its opposite?

If the group is ``all possible permutations'' then of course the answer is yes. If it is the cyclic group then yes again. If it is the dihedral group or either of its sub-groups then yes again. But does there exist a group that is transitive but wouldn't allow for this transformation?


The alternating group $A_6$, acting on $\{e_1, e_2, e_3, a_1, a_2, a_3\}$ in the obvious way is transitive, but the "opposition" transformation is an odd permutation.

  • $\begingroup$ right - so i can get from any element to any other but cant get to $\{a_1,a_2,a_3,e_1,e_2,e_3\}$ because this is an odd permutation and $A_6$ only includes even permutations? Thanks, I don't know group theory! $\endgroup$ – jdizzle Jun 18 '15 at 23:22
  • $\begingroup$ Yup, that's it! $\endgroup$ – Noah Schweber Jun 18 '15 at 23:36

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