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Modern Algebra: Proof of an exponent law

  1. Sep 30, 2014 #1
    The problem is to verify ##(g^n)^{-1} = g^{-n}## is true ##\forall n \in \mathbb{Z}##. Here is my proof:

    ## (g^n)^{-1} = (\underbrace{g \star g~ \star ~...~ \star g}_{n~ \mbox{copies}})^{-1} \iff##

    ##(g^n)^{-1} = [(g \star ~...~ \star g) \star g]^{-1}##

    Using ##(a \star b)^{-1} = b^{-1} \star a^{-1}## (1), which I have already proven, gives

    ##(g^n)^{-1} = g^{-1} \star (\underbrace{g \star g~ \star ~...~ \star g}_{n-1~ \mbox{copies}})^{-1} ##

    If I apply (1) ##n-1## more times, I will arrive at

    ##(g^n)^{-1} = \underbrace{g^{-1} \star ~...~\star g^{-1}}_{n ~\mbox{copies}}##

    The right hand side is by definition ##g^{-n}##.

    What is wrong with this proof? My professor said it is wrong.
     
    Last edited: Sep 30, 2014
  2. jcsd
  3. Sep 30, 2014 #2

    jedishrfu

    Staff: Mentor

    I think the preferred proof would be by induction where you do two steps:
    - prove that when n=1 its true
    - prove given its true for k then its true for k+1
     
  4. Sep 30, 2014 #3

    Mark44

    Staff: Mentor

    Your prof is probably looking for a proof by induction. The "n copies" and "n - 1 copies" stuff is not rigorous.
     
  5. Sep 30, 2014 #4
    Well, the definition includes the word "copy:"

    "In general, if ##n## is a positive integer, then ##g^n## is the binary operation applied to ##n## copies of ##g##."

    I tried using induction; but I am having difficulty using induction to prove the procedure I am using.
     
    Last edited: Sep 30, 2014
  6. Sep 30, 2014 #5
    I think I need to use induction to show that my procedure is valid; the procedure being that I associate a certain amount of elements, and then apply (1). I could denote this procedure by P(n), and then show that P(n) is true for all natural numbers.

    Does this appear correct?
     
  7. Sep 30, 2014 #6

    Mark44

    Staff: Mentor

    Not quite. Along the lines of what jedishrfu said, do these things.
    1. Show that the statement is true for n = 1.
    2. Assume that the statement is true for n = k. IOW, assume that (gk)-1 = g-k.
    3. Show (prove) that the statement is true for n = k + 1. You will need to use the assumption of step 2 to do this.

    BTW, shouldn't the problem read "for all n ##\in## Z+; i.e., the positive integers?
     
  8. Oct 1, 2014 #7

    WWGD

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    Why not just left-multiply by ## g^n ## in the first line?

    Then you get :

    ##(g^n)(g^n)^{-1}=e = g^n g^{-n} ##
     
  9. Oct 1, 2014 #8
    But I am not certain at the moment that ##g^{n}## is the inverse of ##g^{-n}##, as this would rely on the fact that ##g^n g^m = g^{n+m}##, which I am trying to avoid using. However, I could write

    ##g^{n} \star g^{-n} = (g \star g \star ... \star g) \star (g^{-1} \star g^{-1} \star ... \star g^{-1} )##

    From this I could the associative property a finite number of times.
     
  10. Oct 1, 2014 #9

    WWGD

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    No, I am not assuming that ;by def., ##(g^n)^{-1} ## is the inverse of #g^n#
     
  11. Oct 1, 2014 #10

    Mark44

    Staff: Mentor

    The equation above is precisely the type of thing you should use induction on to prove.
     
  12. Oct 1, 2014 #11
    Yes, that is true. But it isn't immediately obvious that ##g^{-n}## is the inverse of ##g^n##.
     
  13. Oct 1, 2014 #12
    Yes, but I need to prove that the statement is true not only for the natural numbers, but also the negative integers. This is what I am having difficulty with.
     
  14. Oct 1, 2014 #13

    WWGD

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    But then you get that the identity 'e' on the left side equals## g^n g^{-n}##. This means precisely that #g^{-n}# is the inverse of #g^n#, since their product equals the identity. a is the inverse of b , by def., if ab=e:=id. then a is the inverse of b and viceversa; ab=id, then left- multiply both sides by ##a^{-1}## to get $$a^{-1}ab=b=a^{-1} $$
     
    Last edited: Oct 1, 2014
  15. Oct 1, 2014 #14

    Mark44

    Staff: Mentor

    Before we go any further, what is g? Is it an element of a group whose operation is multiplication?
     
  16. Oct 1, 2014 #15
    ##g## is a group element with the arbitrary binary operator ##\star##
     
  17. Oct 1, 2014 #16

    Mark44

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    Are there any properties given for this operator?
     
  18. Oct 1, 2014 #17

    WWGD

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    II think this is just a general group operation.
     
  19. Oct 1, 2014 #18
    This is correct.
     
  20. Oct 1, 2014 #19
    But would that not require assuming the equality ##(g^{n})^{-1} = g^{-n}## is true, which is precisely what we are trying to prove?
     
  21. Oct 1, 2014 #20

    Mark44

    Staff: Mentor

    For n a positive integer, you're showing that gn ⋆ g-n = 1 (or whatever the identity is), using induction.
    For m a negative integer, -m is a positive integer, show that g-m ⋆ gm = 1, also by induction. Sharper minds than mine might be able to give a reason to skip this second step.
    If the group is known to be Abelian (commutive), then I don't think the second part above is needed, because a ⋆ b would be the same as b ⋆ a.
     
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