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Homework Help: Proof involving group and subgroup

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  1. Oct 29, 2011 #1

    ArcanaNoir

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    1. The problem statement, all variables and given/known data
    If H is a subgroup of a group G, and a, b [itex]\in[/itex] G, then the following four conditions are equivalent:
    i) [itex] ab^{-1} \in H [/itex]
    ii) [itex] a=hb [/itex] for some [itex] h \in H [/itex]
    iii) [itex] a \in Hb [/itex]
    iv) [itex] Ha=Hb [/itex]


    2. Relevant equations
    cancellation law seems handy, and existence of an inverse, associativity, and other basic things.


    3. The attempt at a solution

    [itex] ab^{-1} \in H \Rightarrow \exists h \in H : ab^{-1}=h \Rightarrow (ab^{-1})b=hb \Rightarrow a(bb^{-1})=ae=a=hb \Rightarrow a \in Hb \Rightarrow ...? [/itex]
     
    ArcanaNoir, Oct 29, 2011
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  3. Oct 29, 2011 #2

    micromass

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    Ok, so you already proved (1)=>(2) and (2)=>(3).

    So assume that (3) holds. So a is in bH.

    You need to prove that aH=bH. So take an element in aH and prove that it is in bH, and take an element in bH and prove that it is in aH. Write these things out.
     
    micromass, Oct 29, 2011
  4. Oct 29, 2011 #3

    ArcanaNoir

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    does H=Hb?
     
    ArcanaNoir, Oct 29, 2011
  5. Oct 29, 2011 #4

    micromass

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    Only if b is in H.

    You need to prove that aH is in bH.
    So take an element ah in aH, you must prove that it is in bH. So you must write it in the form bh'.
     
    micromass, Oct 29, 2011
  6. Oct 29, 2011 #5

    ArcanaNoir

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    I'm stuck. I tried doing something like a=hb but for ha=b but since h=ab^-1 that didn't help, so I considered [itex] h^{-1}=ba^{-1} [/itex] so [itex] ba^{-1}=h^{-1} [/itex] and [itex] h=ab^{-1} \Rightarrow [/itex]
    But I'm not getting anywhere. What should come directly after [itex] a \in Hb \Rightarrow [/itex] ?
     
    ArcanaNoir, Oct 29, 2011
  7. Oct 29, 2011 #6

    micromass

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    Take ah in aH. You must show that it is in bH. So you must find h' in H so that ah=bh'.
    The logical thing to do is to take [itex]h^\prime=b^{-1}ah[/itex]. But why is this in H?
     
    micromass, Oct 29, 2011
  8. Oct 29, 2011 #7

    ArcanaNoir

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    is it a problem that I'm looking for Ha, not aH?
     
    ArcanaNoir, Oct 29, 2011
  9. Oct 29, 2011 #8

    micromass

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    Oh, I'm sorry. I didn't notice that. It doesn't change much.

    You need to find h' such that ha=h'b. So pick [itex]h^\prime=hab^{-1}[/itex]. but why is this in H?
     
    micromass, Oct 29, 2011
  10. Oct 29, 2011 #9

    ArcanaNoir

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    okay, h is in H already, so hh is in H, and [itex] hh=hab^{-1}=h' \in H [/itex]
     
    ArcanaNoir, Oct 29, 2011
  11. Oct 29, 2011 #10

    ArcanaNoir

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    so [itex] h' \in H \Rightarrow h'b \in Hb \Rightarrow ah'b \in Hb \Rightarrow ah' \in H \Rightarrow [/itex]
    [itex] h^{-1}ah' \in H \Rightarrow ba^{-1}ah' \in H \Rightarrow bh' \in H [/itex]
     
    Last edited: Oct 29, 2011
    ArcanaNoir, Oct 29, 2011
  12. Oct 29, 2011 #11

    micromass

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    Why?? This isn't true...
     
    micromass, Oct 29, 2011
  13. Oct 29, 2011 #12

    ArcanaNoir

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    Because [itex] ab^{-1}=h [/itex] so [itex] hh=ab^{-1}h [/itex]
     
    ArcanaNoir, Oct 29, 2011
  14. Oct 29, 2011 #13

    micromass

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    but h isn't [itex]ab^{-1}[/itex].... Where did you get that??
     
    micromass, Oct 29, 2011
  15. Oct 29, 2011 #14

    ArcanaNoir

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    the very beginning, [itex] ab^{-1} \in H \Rightarrow \exists h \in H : ab^{-1}=h [/itex]
     
    ArcanaNoir, Oct 29, 2011
  16. Oct 29, 2011 #15

    micromass

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    But that was a different proof. That was notation in order to prove (1)=> (2). You can't use that now.

    Right now you're proving (3)=> (4). So to do that, you chose [itex]ah\in aH[/itex]. All you know about h now is that it is in H!!
    You're doing something different than in the beginning, so you can't use the notation from previous proofs...
     
    micromass, Oct 29, 2011
  17. Oct 29, 2011 #16

    ArcanaNoir

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    okay I'll pick a different h but I disagree about calling it a "previous proof" since the whole point of this exercise is to make the thing a connected circle.
    so thats why I need something to follow directly from a \in Hb.

    but first it's good to break it down into just 3 => 4
    so, [itex] h \in H \Rightarrow ha \in Ha [/itex]
    Now, choosing [itex] h' : h'=hab^{-1} [/itex] is in H because we assumed [itex] ab^-1 \in H [/itex]
    [itex] ha \in Ha \Rightarrow hab^{-1} \in Hab^{-1} \Rightarrow h' \in Hh [/itex] crap I seem lost again.
     
    ArcanaNoir, Oct 29, 2011
  18. Oct 29, 2011 #17

    ArcanaNoir

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    I'm breaking out the chocolate fudge brownie ice cream now. Are you trying to make me cry? I have a lot of other proofs to study before test Tuesday :(
     
    ArcanaNoir, Oct 29, 2011
  19. Oct 29, 2011 #18

    micromass

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    You need to do 4 proofs here: (1)=> (2), (2)=> (3), (3)=> (4) and (4)=> (1). These are 4 distinct things to prove. You can not prove them simultaniously by one long chain of statements. You need to prove all four things separately.
    Indeed, so h' is in H. So ha=hb' is in Hb, what we needed to prove!!
     
    micromass, Oct 29, 2011
  20. Oct 29, 2011 #19

    micromass

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    I'm sorry, I was only trying to help :frown:

    I'll let somebody else handle the thread then... :frown:
     
    micromass, Oct 29, 2011
  21. Oct 29, 2011 #20

    ArcanaNoir

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    what? Where?
    [itex] ha \in Ha \Rightarrow hab^{-1} \in Hab^{-1} \Rightarrow hab^{-1} \in H [/itex] but how does that show [itex] ha=hb^{-1} [/itex] ?
     
    ArcanaNoir, Oct 29, 2011
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