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Would anyone happen to recall how to put $D_6$ into $S_5?$ The cyclic generator has to go (up to conjugacy) to $(1 2 3) (4 5),$ but where does the involution go?

Jyrki Lahtonen
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Igor Rivin
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    I think this question has a solution. – pjs36 Dec 31 '16 at 21:45
  • I like your answer to that version @pjs36. Pictures are always nice! Do you think that this should be closed as a duplicate? – Jyrki Lahtonen Dec 31 '16 at 21:49
  • Thank you, @JyrkiLahtonen ! Looking back, I realized some of the wording was a bit funky. Anyway, I don't feel strongly one way or the other (at least, I'm not overly inclined to close) -- that question was about a specific isomorphism, to be fair; this one's a bit more broad, I suppose. I think you know better than me! :) – pjs36 Dec 31 '16 at 21:54

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If you try $a\mapsto (123)(45)$, $b\mapsto (23)$, then the key relation $bab^{-1}=a^{-1}$ is satisfied.

Instead of $(23)$ you can use any other 2-cycle acting on $\{1,2,3\}$.

The key to success here is that $D_6\simeq D_3\times C_2$.

Jyrki Lahtonen
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