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let $D_3$ be the dihedral group for a triangle (so it’s the set of all congruences that leave a regular triangle invariant). Consider $S_3$ the set of all permutations of 3 elements. Now my book says that it’s directly obvious that $f\colon D_3\to S_3$ is a homomorphism. I personally don’t see that directly. I’m guessing I should show that: $$ f(d_1\cdot d_2)=f(d_1)f(d_2). $$ So the composition of two congruences should correspond with the composition of its two associated permutations. How can I show that generally?

I first tried out an example. Consider the following two congruences: $c_1=\sigma_{\pi/3}$ and $c_2=\rho_{2\pi/3}$. The corresponding permutations should be $p_1=(12)$ and $p_2=(132)$. However, the problem I stumbled upon is the following:

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As you can see, the permutation that corresponds with $\rho_{2\pi/3}$ seems to depends on whether the triangle is reflected or not. How can that be the case? Shouldn't each permutation be the same for the same rotation, no matter how the vertices 1,2,3 are numbered initially?

I found this problem, because I wanted to check that $$ f(\rho_{2\pi/3}\cdot\sigma_{\pi/3})=f(\rho_{2\pi/3})\cdot f(\sigma_{\pi/3})=(132)\cdot(12)\neq (13), $$ where I found (13) by writing out the trangles:

enter image description here

Sha Vuklia
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  • The isomorphism is not unique, in fact it depends on the chosen identification of the vertices of the triangle with the set ${1, 2, 3}$. Once you fix the identification, then there is a unique well-defined permutation corresponding to each rotation and reflexion. This is clear abstractly: the dihedral group $D_3$ is presented as $$\langle \rho, \sigma , | , \rho^3=\sigma^2=1, , \rho \sigma = \sigma \rho^{-1} \rangle,$$ and there are two possible choices for $f(\rho)$, namely $(1 ,2, 3)$ and $(1, 3 ,2)$, and three possible choices for $f(\sigma),$ namely the three transpositions. – Francesco Polizzi Jun 25 '17 at 13:18
  • Alright, so say we fix the identification, any idea on how to show that $f$ is indeed a homomorphism? – Sha Vuklia Jun 25 '17 at 13:32
  • Checking that your $f$ is a group homomorphism is standard. The image of $\rho$ is a $3$-cycle and the image of $\sigma$ is a $2$-cycle, and together they generate $S_3$, so $f$ is surjective. But the two groups have the same order (namely, six) so $f$ is also injective, hence an isomorphism. – Francesco Polizzi Jun 25 '17 at 13:36
  • Note that this is the only case. For $n \geq 4$, the analogous group homomorphism $f \colon D_n \to S_n$ is only injective but not surjective, since $2n < n!$. Geometrically, this means that there are permutations of the vertices of a regular $n$-gon that do not come from an isometry. – Francesco Polizzi Jun 25 '17 at 13:39
  • @Francesco I'm guessing this standard method you mention is writing out the multiplication table for $S_3$ and $D_3$, and then noting that they are indeed isomorphic? – Sha Vuklia Jun 25 '17 at 14:10
  • If you want, but this is long. To simplify the argument, it suffices to check that your permutations $f(\sigma)$ and $f(\rho)$ satisfy the same relations as $\sigma$ and $\rho$ written in my comment above (this will ensure that f is a group homomorphism) and then deduce surjectivity and injectivity as I did in the other comment. – Francesco Polizzi Jun 25 '17 at 14:17
  • I'm guessing you mean I should check that $f(\rho)^3=f(\sigma)^2=1$, and $f(\rho)f(\sigma)=f(\sigma)f(\rho^{-1})$. I'm not familiar with the theorems that would justify that argument, so for now I'll stick to the Cayley tables, and maybe use this simples approach in the future. – Sha Vuklia Jun 25 '17 at 14:28
  • Exactly. Cayley's tables are useful for groups of small order, checking with them that some map is a group homomorphism for groups of high order is clearly not the best choice. – Francesco Polizzi Jun 25 '17 at 14:33

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