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My question deals with the dihedral group $D_3$ of equilateral triangle 123 (1 top vertex, 2 bottom right vertex, 3 bottom left vertex).

  • R1 is the counterclockwise rotation of 120 degrees.

  • R2 is the counterclockwise rotation of 240 degrees.

  • SA is the symmetry through the top angle bissectrice (exchanging vertex 2 and 3)

  • SB is the symmetry through the right angle bissectrice (exchanging vertex 1 and 3)

  • SC is the symmetry through the left angle bisssectrice (exchanging vertex 1 and 2)

I constructed the Cayley table accordingly, and compared it to the $S_3$ Cayley table: they match very well, hence my deduction that:

  • R1 matches with permutation (123) of $S_3$.

  • R2 matches with permutation (321).

  • SA matches with permutation (23).

  • SB matches with permutation (12).

  • SC matches with permutation (31).

First question, is this correspondence fully right? The shapes of the two tables are quite similar, but I did not try all the possibilities.

Second problem :

I tried to deepen the above correspondence as follows :

If I write a symmetry in a line with 1st figure = top vertex, 2nd figure = right vertex and 3rd figure = left vertex, the symmetry R1 can be written

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One way to prove the isomorphism is the following (this seems to be what your are trying to do in your second question):

The group $S_3$ acts on the set with three points. If you number the vertices of an equilateral triangle 1,2 and 3 then you can see that $S_3$ acts on your triangle. Looking at the action of $S_3$ on your triangle, you can see that you get all of the symmetries of the triangle. As the group of symmetries of the triangle is precisely $D_3$, you are done!$^{\dagger}$

You should realise that this isomorphism is precisely the one you found by looking at the Cayley tables. So your correspondence is correct.


$^{\dagger}$Well, actually this just shows that there is a surjective homomorphism $S_3\rightarrow D_3$. Do you understand why this implies that there is an isomorphism between these groups?

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