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Math Help - Group action

  1. #1
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    Group action

    How can i show a group action where G is the group of symmetries of a regular octagon and X is the set of all possible colourings of a regular octagon where each edge is coloured one of two colours??

    And could someone please provide more examples like this one, may be using a different shape e.g. a square or a triangle?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Becks View Post
    How can i show a group action where G is the group of symmetries of a regular octagon and X is the set of all possible colourings of a regular octagon where each edge is coloured one of two colours??

    And could someone please provide more examples like this one, may be using a different shape e.g. a square or a triangle?
    What do you think? I mean, what have you done in class?
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  3. #3
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    I'm confused by what a group action is. I have the definition but i need some examples, like the one i mentioned above.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Becks View Post
    I'm confused by what a group action is. I have the definition but i need some examples, like the one i mentioned above.
    In my opinion (I guess I'll try to help you with the shape example if the following doesn't work) the most informative examples of group actions are when the groups act on themselves.

    For example, remember Lagrange's theorem? It can be said more "eloquently" in the language of group actions.

    So, let G be a group and H\leqslant G. Then we can define a group action on G by h,g)\mapsto hg" alt="\alpha:H\times G\to Gh,g)\mapsto hg" />. This is a group action since \alpha(e,g)=eg=g,\text{ }\forall g\in G and \alpha\left(h,\alpha(h',g)\right)=\alpha(h,h'g)=hh  'g=\alpha(hh',g). Now, think of the orbits of the relation as \mathcal{O}_g=\left\{x\in G:x=\alpha(h,g)\text{ for some }h\in H\right\}. It is easily shown that that |\mathcal{O}_g|=|H|. Thus, since G=\mathcal{O}_{g_1}\cup\cdots\cup \mathcal{O}_{g_n} and \mathcal{O}_{g_k}\cap\mathcal{O}_{g_\ell}=\varnoth  ing,\text{ }k\ne \ell it follows that |G|=\left|\mathcal{O}_{g_1}\cup\cdots\cup\mathcal{  O}_{g_n}\right|=n|H| which implies that |H|\mid |G|
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  5. #5
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    Sorry, but that is too advanced for me or so it seems, and would it be possible for you to use simpler notation or english to explain things a bit more. Thanks for the above though
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