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Isomorphic groups G and H, G has subgroup order n implies H has subgroup order n

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data

    G is isomorphic to H.
    Prove that if G has a subgroup of order n, H has a subgroup of order n.

    2. Relevant equations

    G is isomorphic to H means there is an operation preserving bijection from G to H.

    3. The attempt at a solution

    I don't know if this is the right approach, but
    G has a subgroup of order n implies e is in the subgroup, and since e in G maps to e in H, e is in the potential new subgroup of H.
    Suppose g is an element in the subgroup of G. Then, g maps to some h in the new subgroup of H.

    Man, I think I'm proving the wrong thing here. I'm heading toward proving a subgroup in H, not that the order is the same. I need a new direction. Any one have a direction idea?
     
  2. jcsd
  3. Nov 13, 2011 #2

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    Start with the exact definition of a subgroup.
    What is it?

    Verify that the subgroup-conditions check out on the mapping of the subgroup.
     
  4. Nov 13, 2011 #3
    A subgroup is a subset of a group which is closed on the group's operation, the identity in the group is in the subgroup, and each element in the subgroup's inverse is in the subgroup too.

    Let the subgroup of G be g and the potential subgroup of H be h.
    Since e is in any subgroup of G, [tex] \theta (e_g) = e_h [/tex] is in h.
    [tex] \forall x \in g, \exists y \in h : \theta (x) = y [/tex] (because theta is bijection)
    Since x is in g, x-1 is in g.
    How do I show that this means y-1 is in h?
    Also, how do I show a*b is in g implies c*d is in h?
     
  5. Nov 13, 2011 #4

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    Use [itex]\theta(a*b)=\theta(a)*\theta(b)[/itex].
     
  6. Nov 13, 2011 #5
    Does [itex] a*b \in g \Rightarrow \theta (a*b) \in h [/itex] ?
    Or perhaps [itex] a, b \in g \Rightarrow \theta (a) \wedge \theta (b) \in h [/itex]?
     
  7. Nov 13, 2011 #6

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    Let's define h to be θ(g).

    So if a*b is in g, then θ(a*b) is in h!

    We know that g is closed, contains e, and contains the inverse of each of its elements.

    Does h?
     
  8. Nov 13, 2011 #7
    Ah, the critical piece. Sometimes I lack a most basic but most critical piece. *computing*

    θ(g) is certainly a better name for it than the "potential subgroup h"...
     
  9. Nov 13, 2011 #8
    [tex] a,b \in g \Rightarrow a*b \in g \Rightarrow \theta (a*b) \in h \Rightarrow \theta (a)*\theta (b) \in h [/tex] is this enough to show h is closed? (oh, and pretend like I'm using the h operation for h.. I don't feel like going back and changing to a new symbol at this point)

    [tex] a \in g \Rightarrow a^{-1} \in g \Rightarrow \theta (a^{-1}) \in h \Rightarrow (\theta (a))^{-1} \in h [/tex]

    Then, of course the number of elements is the same because theta is a bijection. So, All proved, right?
     
  10. Nov 13, 2011 #9

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    Hmm, it's sort of proved... but it's a bit sloppy and on an exam they like rigorous proofs...



    To show h is closed, you need to start with the definition of closure, which is:
    For each c and d in h, c*d is also in h.​


    That is, you need to start with:
    For each c and d in h, ...​
    and you need to end with:
    ..., so c*d is in h.​



    Same thing for the inverse. Start with:
    For each [itex]a[/itex] in h, ...​
    and end with:
    ..., so there is a [itex]b[/itex] in h with [itex]a*b = b*a = e[/itex].​



    That is, assuming you want a perfect grade! :wink:
     
  11. Nov 13, 2011 #10
    Well, this is only studying so perfection isn't needed all the time. So long as I know what the idea is. I think I'm close enough on this one to move on comfortably.

    One time I made the mistake of asking my prof about a homework problem, and showing him my homework pages (which we did not turn in). It was the tenth page of induction proofs, so I had stopped putting "assume this for k, that is assume (some equation) is true". He chastised me for my un-thoroughness. There just comes a point when ten pages into something I become un-thorough. :P

    Plus, Latex takes so long to write. I think someday I'll get one of those fancy keyboards that you can program a bazillion functions into the buttons, and i'll make latex buttons.
     
  12. Nov 13, 2011 #11

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    I get it. You don't like to type.


    I'd like to make another point though.
    If you are wondering in a problem what it is exactly what you need to do, you usually need to drag up the exact definitions, and simply repeat them.
    Then repeat them again, substituting different symbols into them as apply to the problem at hand.

    From there, the solution often becomes suddenly obvious.


    Btw, didn't you post in some thread this nifty tool that converts handwriting into latex? Or am I mixing this up? :wink:
     
  13. Nov 13, 2011 #12
    Handwriting into latex? Nope, I didn't post anything like that.

    The real problem with this section is our professor suddenly realized the semester is almost over and we have two tests to do yet, so he kind of rushed us on this section. He left out a lot of the types of problems I'm working on now. I'm an example-based learner.

    Anyway, thanks for the help, as always :)
     
  14. Nov 13, 2011 #13

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