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For me $D_{6}$ is the dihedral group of $2n$ elements as on Wikipedia.

the stabilizer of $\{2,4,6\}$ in $D_{6}$ is defined here Understanding how reflection acts on rotation in $D_{6}.$ which is the following set $\{id. , \rho^{2}, \rho^{4}$, reflection through the axis of symmetry passing through the vertex 2 and between vertices 4 & 6, reflection through the axis of symmetry passing through the vertex 4 and between vertices 2 & 6, reflection through the axis of symmetry passing through the vertex 6 and between vertices 2 & 4 } where $\rho$ means rotation.

My trial to create a homomorphism is this:\

$() \rightarrow id., (12) \rightarrow \rho^{2}, (13) \rightarrow \rho^{4}, (23) \rightarrow $ reflection through the axis of symmetry passing through the vertex 2 and between vertices 4 & 6, $(1 2 3) \rightarrow $ reflection through the axis of symmetry passing through the vertex 4 and between vertices 2 & 6 and $(132) \rightarrow $ reflection through the axis of symmetry passing through the vertex 6 and between vertices 2 & 4 .

Is this a correct isomorphism? Is there a smarter way of saying that they are isomorphic?

1 Answers1

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You do not need to construct an explicit isomorphism.

  1. Compute the number of elements in the stabilizer. It must be $6$.

  2. Prove that the stabilizer is not commutative.

  3. Use the fact that there is exactly one non-commutative group of order $6$, the group $S_3$.

markvs
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  • Is there a rigorous way of proving that the stabilizer is non-commutative? –  Oct 06 '20 at 03:02
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    One approach would be to note that it has subgroups of order $2$ and $3$ by Cauchy's theorem, and then show that you have $\Bbb Z_3\rtimes\Bbb Z_2$. –  Oct 06 '20 at 03:13
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    @Math: The most rigorous way is to just find two elements there that do not commute. I would take any two reflections in the stabilizer. – markvs Oct 06 '20 at 03:15
  • Sorry, I do not understand your first comment answers which question of mine. –  Oct 06 '20 at 08:37