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Question about grp of order 6

  1. Jul 18, 2011 #1
    Lemma. A group [itex]G[/itex] of order 6 can have only one element of order 3.

    Pf. Suppose [itex]G[/itex] has two elements of order 3. Call these elements [itex]x[/itex] and [itex]y[/itex]. Let [itex]H[/itex] and [itex]K[/itex] be the subgroups generated by [itex]x[/itex] and [itex]y[/itex] resp. Then [itex]H \cap K = \{ e \}[/itex] and therefore [itex]G[/itex] can have only one subgroup of order 3.

    I'm reading over my notes from class and I'm confused on the reasoning here. Why does [itex]H \cap K = \{ e \}[/itex] imply that [itex]G[/itex] can have only one element of order 3?
  2. jcsd
  3. Jul 18, 2011 #2
    How many elements does

    [tex]HK=\{hk~\vert~h\in H,k\in K\}[/tex]

    have if [itex]H\cap K=\{e\}[/itex]. That is, can you find more than 6 elements in this group?
  4. Jul 18, 2011 #3
    Hi micromass.

    Let [itex]H = \{ e, x, x^2 \}[/itex], and let [itex]K = \{ e, y, y^2 \}[/itex]. If [itex]H \cap K = \{ e \}[/itex], then [itex]HK = \{ e, xy, x^2y, x^2y, x^2y^2, yx, yx^2, y^2x, y^2x^2 \}[/itex] and [itex]|HK| = 9[/itex]. And since [itex]e, x, x^2, y, y^2[/itex] are unique, therefore there products constitute unique elements. Thus [itex]G[/itex] can have only one element of order 3.
  5. Jul 18, 2011 #4
    No, that's not correct. Why can't xy=yx, for example?? Furthermore, you forgot x,y,x2 and y2.

    Let's consider this set:


    Can you show that all of these elements are distinct from each other??
  6. Jul 18, 2011 #5
    Bummer. That was my wishful reasoning there. Well by cancellation laws, I can determine that:

    1. [itex]x \neq x^2[/itex] since if they were equal, this would imply [itex]x=e[/itex], but [itex]x[/itex] has order 3.
    2. [itex]x \neq y[/itex], by assumption.
    3. [itex]x \neq xy[/itex] since [itex]y \neq e[/itex].
    4. [itex]x \neq x^2 y[/itex] since if they were equal, by cancellation laws, it would follow that [itex]x^{-1} = x^2 = y[/itex], which can't happen since [itex]x[/itex] and [itex]y[/itex] generate unique elements.
    5. [itex]x \neq y^2 x y^2 x^2[/itex] since if they were equal, it would follow that [itex]y=e[/itex].
    6. [itex]x \neq y^2[/itex] since [itex]x[/itex] and [itex]y[/itex] generate unique elements.

    A argument that the other elements are not equal is similar.
  7. Jul 18, 2011 #6
    Uuuh, ok, that's good. But you can do things a lot easier. We just need to prove that

    [tex]hk\neq h^\prime k^\prime[/tex]

    for h,h' in H and k,k' in K. But if equality holds then

    [tex]h^\prime h =k^\prime k\in H\cap K[/tex]

    that's what your professor wants you to do.
  8. Jul 18, 2011 #7
    Okay, I think I see how to do this. Suppose [itex]H \cap K = \{ e \}[/itex]. I want to suppose that there exists [itex]h, h' \in H[/itex] and [itex]k, k' \in K[/itex] such that [itex]hk = h'k'[/itex]. It follows that [itex]h^{-1}h' = kk'^{-1}[/itex]. Let [itex]h'' = h^{-1}[/itex] and [itex]k'' = k'^{-1}[/itex] and get [itex]h'' h' = k k''[/itex]. But this would imply both [itex]h'' h[/itex] and [itex]k k''[/itex] belong to [itex]H \cap K[/itex], a contradiction. How does this look?
  9. Jul 18, 2011 #8
    Why is this a contradiction? Why can't both h''h and k'' be in [itex]H\cap K[/itex]?
  10. Jul 18, 2011 #9
    I was thinking that it was contradiction to my assumption that [itex]H \cap K = \{ e \}[/itex], but now I realize that [itex]h^{-1} h'[/itex] could equal the identity. But it seems strange to me that for different elements [itex]h, h', k, k'[/itex] that if [itex]hk = h'k'[/itex] then [itex]h h'^{-1}[/itex] could be the identity because wouldn't that be saying h=h' and k=k'? That is why I thought that if I showed that [itex]hk=h'k'[/itex] implied [itex]h'' h'[/itex] and [itex]kk''[/itex] belonged to [itex]H \cap K[/itex], it would contradict my supposition since I was assuming [itex]h'' h' \neq e[/itex].
    Last edited: Jul 18, 2011
  11. Jul 18, 2011 #10
    Yes, indeed, so if hk=h'k', then you have h=h' and k=k'. Wasn't this what you needed to show?
  12. Jul 18, 2011 #11
    So what you're saying is instead of citing a contradiction, what I should of concluded from my proof above that since [itex]h'h''[/itex] and [itex]kk''[/itex] belong to [itex]H \cap K[/itex], it follows that [itex]h'h'' = e = kk''[/itex] since [itex]H \cap K = \{ e \}[/itex]. Thus [itex]h=h'[/itex] and [itex]k=k'[/itex].
  13. Jul 18, 2011 #12
  14. Jul 18, 2011 #13
    Thank you very much for all your help, micromass!
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