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


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?

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.