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Homework Help: Find all subgroups of the octic group

  1. Aug 2, 2011 #1
    Here is my work thus far, and I included any pertinent notes.

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110802_182239.jpg?t=1312327545 [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 2, 2011 #2
    Looks good so far, shackleford! :smile:


    I'll also assume that the "octic group" is just the dihedral group of order 8. I'm not familiar with the name octic group...
     
  4. Aug 2, 2011 #3
    It's the operations on the square numbered

    4 3
    1 2.

    I'm missing an element from the subgroup of order 4.
     
  5. Aug 2, 2011 #4
    Indeed, that would be the dihedral group. :smile:

    Well, what happens if you do (1 2 3 4)(1 2 3 4)?? Which familiar element do you get?
     
  6. Aug 2, 2011 #5
    Well, I figured I could simply do another permutation different than alpha and alpha-squared.

    (1,3,2,4)
     
  7. Aug 3, 2011 #6
    (1 2 3 4)(1 2 3 4) = (1,3)(2,4) = α2

    I already have (1,3)(2,4) in the subgroup of order 2.
     
  8. Aug 3, 2011 #7
    Your elements can occur in multiple subgroups. That's not forbidden...
     
  9. Aug 3, 2011 #8
    I understand that, but the order is not four. The order of the permutation is the LCM of the order of the individual cycles.
     
  10. Aug 3, 2011 #9
    Also, the next problem is Find all normal subgroups of the octic group.

    Wouldn't that be all the subgroups since they each have the identity in them? The left and right cosets would be equal since the identity captures all of the elements in the octic group.
     
  11. Aug 3, 2011 #10
    Yes, the order of a2 is 2. But why would that prevent {1,a,a2,a3} to form a subgroup??
     
  12. Aug 3, 2011 #11
    What does having the identity has to do with normality??

    Calculate the left and right cosets of [itex]\beta[/itex] for example. You'll see that the left and right cosets will not be equal...
     
  13. Aug 3, 2011 #12
    Oh, you're right. The order of that element divides 4. In general then, the subgroup is not unique then because I could stick another second-order element in its place.
     
  14. Aug 3, 2011 #13
    Look at my subgroups. Each of the subgroups have the identity.

    Normality is set equality.

    xH = Hx for all x in G.

    H = G1,...,G8 = G.
     
  15. Aug 3, 2011 #14
    No, you couldn't :biggrin:

    Having a in the subgroup will force a2 in the subgroup, by definition.
     
  16. Aug 3, 2011 #15
    Yes, all subgroups of every group will contain the identity. This has nothing to do with normality.
     
  17. Aug 3, 2011 #16
    Oh. The subgroup has to be closed. That's why alpha-squared has to be in there.
     
  18. Aug 3, 2011 #17
    Duh. Sorry. I had a brain fart there. I need to look at the product of each subgroup with each element in G to determine if the subgroup is normal in G.

    I have the handy group table in the back of the book.
     
  19. Aug 3, 2011 #18
    Okay. I found that the only two normal subgroups of the octic group are, according to my notation, G1 = {e} and G2 = {e, alpha2}. G is not normal in G.
     
  20. Aug 3, 2011 #19
    These aren't all the normal subgroups yet, you're missing one. (Hint: subgroups of index 2 are always normal)

    By the way, I'm also not convinced that you actually found all the subgroups either...
     
  21. Aug 3, 2011 #20
    Then I have no idea how to find them all. How do I find all subgroups of the octic group?
     
    Last edited: Aug 3, 2011
  22. Aug 3, 2011 #21
    Well, first you take an arbitrary element x and see what the subgroup generated by x is. This will give you all cyclic subgroups of the octic group. This is what you have done.

    But you're not done yet. There-after, you will have to take two arbitrary elements x and y, and you'll have to see what kind of group that those two elements generate.

    Then you'll have to take 3 elements and see what they generate. And so on...

    This sounds like a lot of work, but it isn't. If you're smart, then you can cut a lot of work.

    For example, let's say that you are in STEP 2 and you see what group is generated by 2 elements. Obviously, you don't need to check it for a and a2, since these two elements will lie in the same cyclic subgroup. And obviously, the group generated by a and delta is the same as the group generated by a3 and delta.
     
  23. Aug 3, 2011 #22
    It also looks like G7 is a normal subgroup, too.
     
  24. Aug 3, 2011 #23
    Okay. I generated the subgroups based on the distinct powers of the elements of G. The only nontrivial generated subgroups are

    <α> = {α, α2, α3, e}
    3> = {α, α2, α3, e}

    I'm not smart. I'm not following your shortcut.

    Do I multiply α, α2, α3 by each of the other elements? I see they did that for A = {alpha, beta}.

    If I do alpha with beta, gamma, theta, and delta I get {gamma, delta, beta, and theta}. It looks like I would get similar results with alpha squared and cubed.
     
    Last edited: Aug 4, 2011
  25. Aug 4, 2011 #24
    Okay, I found that someone had already worked out this problem in response to a question on a different website.

    If you adjoin any element with <a> or <a3> you get G. How do you quickly see this? Just by looking at the table?

    Adjoining the other elements with <a2> gives you two distinct subgroups of order 4. I suppose I'm missing another subgroup of order 4. Well, now, I have these two new subgroups, the identity, but it would appear I'm missing one.

    http://i111.photobucket.com/albums/n149/camarolt4z28/20110804161007509.jpg?t=1312492825 [Broken]
     
    Last edited by a moderator: May 5, 2017
  26. Aug 4, 2011 #25
    <a> has order 4, adjoining an element to <a> gives you a subgroup with order at least 5. But by Lagrange's theorem, the order must divide 8. So adjoining an element to <a> gives you a group of order 8: the entire group.

    [/QUOTE]
    Adjoining the other elements with <a2> gives you two distinct subgroups of order 4. I suppose I'm missing another subgroup of order 4. Well, now, I have these two new subgroups, the identity, but it would appear I'm missing one.

    http://i111.photobucket.com/albums/n149/camarolt4z28/20110804161007509.jpg?t=1312492825 [Broken][/QUOTE]

    Why do you get the impression that you're missing one?
     
    Last edited by a moderator: May 5, 2017
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