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Find all subsets of the octic group.

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data
    G = {e, a, a^2, a^3, b, y, D (delta), T (theta)} Where e=(1), a=(1, 2, 3, 4), a^2 = (1, 3)(2, 4), a^3 = (1, 4, 3, 2), b = (1, 4)(2, 3), y = (2, 4), D = (1, 2)(3, 4), T = (1, 3)

    Find the subsets.


    2. Relevant equations

    I know that the order of G is 8. So, my subsets should have an order of 1, 2, 4, or 8

    3. The attempt at a solution

    I am looking at other examples and I do not understand how they make the subgroups. Is it just guessing or is there a method.

    By googling, I know that there are 10 subgroups. So my subgroups with order 1 are just {e}, subgroups with order 2 are {e, a^2} {e, b} {e , y} {e, D}, {e, T}

    The next ones are where I get confused. I would think that my subgroups with order 4 would be {e, a} and {e, a^3} but internet resources say that it is {e, a, a^3} but why?

    And, using {e, a, a^3}, i only get 8 subgroups (including G). What are the other two?

    I suppose my general question is: is there a method?
    Thanks!
     
  2. jcsd
  3. Aug 4, 2011 #2

    micromass

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    Hmm, you're the second one on this forum asking the exact same question...

    Anyway, some remarks:
    1) You probably mean subgroup instead of subset.
    2) Neither {e, a}, {e,a3} or {e,a,a3} is a subgroup. The reason is that subgroups must be closed under the operation. So if a is in the subgroup, then so must a*a.
    3) There is a method. First find all cyclic subgroups, then find the subgroups generated by two elements, then find the subgroups generated by three elements, and so on. This seems to be a lot of work, but it isn't if you're a bit smart.
     
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