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Finding the inverse of a group

  1. Aug 25, 2007 #1
    1. The problem statement, all variables and given/known data

    If a(sub_1), a(sub_2), ...,a(sub_n) belong to a group, what is the inverse of a(sub_1)*a(sub_2)*...*a(sub_n)


    2. Relevant equations

    no equations are used

    3. The attempt at a solution

    Suppose b and c are both inverses of a(sub_1), a(sub_2), ...,a(sub_n). Then a(sub_1)*a(sub_2)*...*a(sub_n) *b =e and a(sub_1)*a(sub_2)*...*a(sub_n) * c=e

    Therefore a(sub_1)*a(sub_2)*...*a(sub_n) *b= a(sub_1)*a(sub_2)*...*a(sub_n) *c

    Through cancellation b=c.

    I think I only proved that a(sub_1)*a(sub_2)*...*a(sub_n) has an inverse. I didn't really find the inverse of a(sub_1)*a(sub_2)*...*a(sub_n)

    How would find the inverse?
     
  2. jcsd
  3. Aug 25, 2007 #2
    Suppose a and b are in a group G, what is the inverse of ab?
     
  4. Aug 25, 2007 #3
    Thats easy. the inverse of ab would just be ba.
     
  5. Aug 25, 2007 #4
    Is it?

    So you are saying that (ab)(ba)=abba=ab2a=e in general for any group?
     
  6. Aug 25, 2007 #5

    quasar987

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    The fact that a(sub_1)*a(sub_2)*...*a(sub_n) has an inverse is implied in the axioms of a group. One of the axioms is "To every element a in the group, there is an inverse a^-1 such that aa^-1=a^-1a=e", remember? What you proved is that the inverse is unique, not that is exists. More generally,

    Let b and c be inverses to a. Then ab=e ==>c(ab)=ce ==>(ca)b=c ==>eb=c ==>b=c.

    I supposed that the inverse of a exists and that b and c are such inverses. I came out of the chain of implication with b=c. So the conclusion is "Supposing the inverse exists, it is unique". there is nothing in my reasoning (which is the same as yours) that implies the existence of the inverse. But thankfully, it is part of the definition of a group that every element has an inverse.


    Also, I would like to know if you yourself know what it means to "cancel", as in when you say

    Consider the equation ab=ac. Then to "cancel" the a on both sides means to multiply both sides of the equality from the right by the inverse of a:

    ab=ac ==> (a^-1)ab=(a^-1)ac ==> (a^-1a)b=(a^-1a)c ==> eb=ec ==> b=c

    You thought you had proved the existence of the inverse but you used the fact that it exists when you "canceled" on both sides.

    These are the flaws I found in your reasoning. I hope you will benefit from the criticism.


    Now for a hint to the actual problem: Use the fact that the inverse of each a(sub_i) exists to construct the inverse of a(sub_1)*a(sub_2)*...*a(sub_n).
     
    Last edited: Aug 25, 2007
  7. Aug 25, 2007 #6
    I was referring to the rule for an inverse: ab=ba=e
     
  8. Aug 25, 2007 #7
    Alright, but I never said that a, and b were inverses of one another. I asked that if a and b were two arbitrary elements in a group, then what is the inverse of their product ab? Or at least that is what I meant to ask.
     
  9. Aug 25, 2007 #8
    Excuse my ignorance, but I don't understand what the a(sub_i) variable represents in relation to a(sub_1)*a(sub_2)*...*a(sub_n)
     
  10. Aug 26, 2007 #9

    matt grime

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    It isn't a variable. It is an element of the group. You have an arbitray set of n elements

    [tex]a_1,\ldots,a_n[/tex]

    and you're asked to find its inverse, which is easy to do in terms of the inverses of the elements [itex]a_i[/tex], i=1,..,n.
     
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