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Homework Help: Abstract Algebra!Desperately in need for help?

  1. Oct 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that the octic group [tex] D_4[/tex] has no subgroups of order, 3, 5, 6 and 7.

    I would appreciate any help on this one.
    Thanx in advance!

    2. Relevant equations

    3. The attempt at a solution

    I usually have at least an idea on how to start about proving things, but lol.. about this one i really have no clue how to start.

    I think i have to use proof by contradicton, but really don't know how to go about it.
    I read somewhere that the octic group, since it is of order 8 it has only subgroups of the order of the factors of 8, that is 1,2,4 and 8, but there was not much about it, so i didn't understand a damn thing, there wasn't any proof about it or any further explanation. I also looked at my book, it seems like i can find nothing that would relate somehow to proving this. We haven't worked that much with the order of groups, we havent done yet the lagrange theorem, or how's it called.
    Last edited: Oct 7, 2008
  2. jcsd
  3. Oct 7, 2008 #2


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    The proof of Lagrange's theorem is not hard. I encourage you to take a look at it because you will learn it sooner or later.

    See which subgroups are generated by the 8 elements (or if you have learned about generators, by the 2 generators) of [itex]D_n[/itex] and argue that any subgroup is a union of those.
  4. Oct 7, 2008 #3
    I believe your hint was extremely helpful, i think i managed to get the job done, however i am gonna post all my work later at some point, if you could have a look at it. I need to run now!

  5. Oct 9, 2008 #4
    I'm back!
    here is how i went about proving it, trying to use CompuChip's hint.

    the octic group is:
    [tex]D_4=\{ (1),(1234),(13)(24),(1432),(24),(14)(23),(13),(12)(34)\}[/tex]

    for convinience let [tex]\rho =(1234), \theta=(24)[/tex]

    So, what i did is, i found all cyclic subgroups of the octic group, namely i found all subgroups generated by the 8 el of [tex] D_4[/tex], that is i found
    [tex][(1)],[\rho],[\rho^2],[\rho^3],[\rho \theta],[\rho^2 \theta],[\rho^3 \theta][/tex]

    I also found two non-cyclic groups, that is

    [tex] K=\{(1),(13),(24),(13)(24)\}, M=\{(1),(12)(34),(13)(24),(14)(23)\}[/tex]

    So, as we can see, all the above subgroups are of orders, 1, 2 or 4, so in order to find subgroups of order 3,5,6, or 7 i claimed that we need to take the union of any of these subgropus. In more details here it is how i proceeded:

    I said, let A and B, be any of [tex]\{[(1)],[\rho],[\rho^2],[\rho^3],[\rho \theta],[\rho^2 \theta],[\rho^3 \theta],K,M\}[/tex] or the union of any of the elements in this set, such that, neither [tex] A\subset B, nor, B\subset A[/tex]

    Now, my claim was that [tex] A\cup B[/tex] does not form a subgroup in [tex]D_4[/tex] thus we cannot have subgropubs of order 3,5,6 or 7 in it.


    I used proof by contradiction, i said, let's suppose that [tex] A\cup B[/tex] is a subgroup in [tex] D_4[/tex]. Thus, it means that it is also closed under the opertaion of [tex] D_4[/tex].

    Now, let [tex] a\in A \setminus B[/tex] and [tex] b\in B \setminus A[/tex] I took these two elements this way, because i wanted to increase the nr. of elements when we would take the union of A and B. So, since AUB is a subgroup it means that

    [tex]ab=h\in A \cup B[/tex] but since [tex] b=a^{-1}h[/tex] this cannot be possible, since it would mean that

    [tex] b=a^{-1}h\in A or b=a^{-1}h\in B[/tex] but this is not possible, since we have supposed that [tex] b\in B \setminus A[/tex].

    So, this means that [tex] A \cup B[/tex] is not a subgroup in [tex] D_4[/tex] and thus we cannot have subgroups of order 3, 5, 6 or 7.

    So, is this even close to being the right way of proving it?

    P.S. Now i know how to prove this one using Lagrange's theorem, i learned it.
  6. Oct 10, 2008 #5
  7. Oct 10, 2008 #6


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    What's all the fuss about? D4 has order 8. Lagrange's theorem says H is a subgroup of D4, the order of H divides 8. 3, 5, 6 and 7 don't divide 8. So there is no subgroup of those orders. Case closed. Done.
  8. Oct 10, 2008 #7
    I know that it is a piece of cake proving it using Lagrange's theorem. But this problem was in the section previous to the chapter where the Lagrange's theorem was first introduced, so the book said that we are supposed to prove it without using the Lagrange's theorem.
  9. Oct 10, 2008 #8


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    Ok, then didn't you say that you had found all of the subgroups of D4 just by calculating? I don't know what unioning these subgroups buys you.
  10. Oct 10, 2008 #9
    Pardone my ignorance, but, I am not sure i am completely getting your point here.
    yeah, i found some subgroups, as stated in my post, of D_4, by using its generators. And also by observation i found those other two non-cyclic subgroups. SO, then i thought that every other subgroup of D_4 should be obtained by taking the union of these subgroups generated by its elements. And then, as i said, i proved that this union is not a subgroup in D_4, so there won't be any subgroups of order 3,5,6 or7.
  11. Oct 10, 2008 #10


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    I didn't say you were ignorant. The union of two distinct subgroups that aren't subsets of each other is never a subgroup. I think that's what you actually proved. I think the point to the exercise, if you haven't done Lagrange's theorem, is just to figure out all of the subgroups of D4 and observe that they all are order 1,2,4 or 8. Just because the union of two subgroups has order 3, doesn't mean there is NO subgroup of order 3.
  12. Oct 10, 2008 #11
    Well, yeah, as you saw, i figured out 10 subgroups of [tex] D_4[/tex] but how can i be sure that this is the max nr. of subgroups in D_4. I mean how could i be sure that these are all the subgroups of D_4, and there are no other subgroups. That's why, i kinda followed comuchip's advice, and claimed that there are other subgroups that are created by taking the union of these subgroups, and derived that contradiction.

    SO, is there any way to figure out that we have found all possible subgroups of a group? I mean, without making use of Lagrange's theorem, cuz if the latter is the case, then i think it is not that difficult to do so.??

  13. Oct 10, 2008 #12


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    I think what compuchip was saying is to think of taking all of the subsets of the elements of the group and figure out what subgroup they generate. This isn't that bad since the order of the group is small and there are only two generators, rho and theta.
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