Find injective homomorphism from D_2n to S_n

  • Thread starter Mr Davis 97
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In summary: Edit:So would this symmetry in general look like ##(1~~n-1)(2~~n-2)\dots (n/2-1~~n/2+1)?And would the odd case look like ##(1~~n-1)(2~~n-2)\dots...(n/2-1~~n/2+1)?
  • #36
I think we went overkill on detail. For the purpose of the problem, you could probably just say: labeling the vertices associates to each element of [itex]D_{2n}[/itex] a permutation in [itex]S_n[/itex]. This association is a homomorphism because multiplication in [itex]D_{2n}[/itex] corresponds to composing permutations. This association is injective because only the identity element of [itex]D_{2n}[/itex] fixes all vertices. Then write down the map explicitly as we did.
 
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  • #37
Infrared said:
I think we went overkill on detail. For the purpose of the problem, you could probably just say: labeling the vertices associates to each element of [itex]D_{2n}[/itex] a permutation in [itex]S_n[/itex]. This association is a homomorphism because multiplication in [itex]D_{2n}[/itex] corresponds to composing permutations. This association is injective because only the identity element of [itex]D_{2n}[/itex] fixes all vertices. Then write down the map explicitly as we did.
You're probably right, and I'll take that into account when writing my solution. There's one more thing I'm curious about. My problem says "Find, with justification, an injective group homomorphism from ##D_{2n}## to ##S_n##." This makes it seem like I have to find one that works for all cases of ##n##. But I had to split it into two cases to get two different homomorphisms... Is the problem just phrased poorly?
 

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