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Abelian group where a*a=e

  1. Feb 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Let G be a group such that a*a = e for all a[tex]\in[/tex]G. Show that G is abelian.

    2. Relevant equations



    3. The attempt at a solution
    I know the conditions for an abelian group but don't see or understand the significance that a*a=e here?
     
  2. jcsd
  3. Feb 11, 2008 #2

    StatusX

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    A group is abelian iff [a,b]=aba-1b-1 is equal to the identity for all a,b (do you see why?). What can you say about a-1 and b-1?
     
  4. Feb 11, 2008 #3
    Yes because a*a[tex]^{}-1[/tex] = e and b*b[tex]^{}-1[/tex] = e. Can you say that the inverses are multiplicatively commutative. But why does the question talk about a*a=e. I would have thought it was a typo or something because it seems strange?
     
  5. Feb 11, 2008 #4

    morphism

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    StatusX gave you a very good hint: G is abelian iff aba-1b-1=e for any a,b in G. The condition a2=e tells you precisely what a-1 is.
     
  6. Feb 11, 2008 #5
    Does that make a^-1 = e^-0.5 ?
     
  7. Feb 11, 2008 #6
    Or could it be that a and its inverse are equal?? *confused*
     
  8. Feb 11, 2008 #7
    Im not sure i do see why, unless what i did write was correct that a*a-1=e and the same for b.
     
    Last edited: Feb 11, 2008
  9. Feb 11, 2008 #8
    Oh does it show it is abelian because for it to equal e, aba-1b-1 would need to also equal aa-1bb-1 which shows the operation is commutative.

    If im still not on the right tracks, i assume that i need to show commutativity anyhow.
     
  10. Feb 11, 2008 #9
    I have to hand in my solutions in half an hour, so one last go at this if anyone can help. I need to show that G is abelian and it says a*a=e. To show its abelian i assume i have to demonstrate the elements are commutative. I guess i need to show elements only of a as no others are mentioned. So a=a^-1.

    a*a^-1 = a^-1*a=e therefore commutative and abelian.
     
    Last edited: Feb 11, 2008
  11. Feb 11, 2008 #10

    morphism

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    A group G is abelian if and only if ab=ba for any a and b in G. This is equivalent to saying that aba-1b-1=e for any a and b in G (why?). Now in your group, a2=e, or equivalently, a*a=e for all a. Thus, because inverses are unique (why?), a-1=a. So now we see that the expression aba-1b-1 is equal to abab (why?). And finally: why is this equal to e?
     
  12. Feb 11, 2008 #11
    Because e=1 so e*e = e?
     
  13. Feb 11, 2008 #12

    morphism

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    No, that's not right. Why do you say that?
     
  14. Feb 11, 2008 #13
    I'm very new to this, ive missed a whole semester on this stuff, but need to do it to do further modules of group theory. Oh well, i will try harder next time.
     
  15. Feb 11, 2008 #14

    morphism

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    That's a good idea. You need to put in some effort to understand this stuff. Reading your posts gave me the impression that you really don't understand the fundamental ideas at work here. Better luck next time I suppose. :smile:
     
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