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Group question

  1. Aug 24, 2007 #1
    1. The problem statement, all variables and given/known data

    In a group, prove that (a^-1)^-1

    2. Relevant equations

    no equations required

    3. The attempt at a solution

    the inverse of a is 1/a and the inverse of 1/a is a. therefore , (a^-1)^-1 = a for all a.

    Also for the property of an exponent, (a^n)^m=a^(m*n)
    so , (a^-1)^-1=a^(-1*-1)=a,

    There for (a^-1)^-1 = a for all a

    Did I use the right methods for proving that (a^-1)^-1 =a?
  2. jcsd
  3. Aug 24, 2007 #2


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    No, no, the exponent method is not okay. It has nothing to do with this problem in group theory. The exponent "-1" is just a shorthand notation for the inverse in a group, it's not the exponent one meets when working with numbers, real, complex, etc.

    So let's see

    Do you agree that

    [tex] \left(a^{-1}\right)^{-1} \cdot a^{-1} = e [/tex] ??

    "e" is the identity in the group.

    If so, if you agree, then consider multiplying the whole relation to the right by "a". See what you get.
    Last edited: Aug 24, 2007
  4. Aug 24, 2007 #3
    This method seems circular. Here are the results of my new proof:

    (a^-1)^-1 * a^-1 -e => a^(-1*-1) *a^(-1)=a^0 =1 =e

    by multiplying a on both sides I get
    a*((a^(-1))^-1 *a^-1) = a*e

    a*(a^(-1*-1) *a^-1)=e*a
    a's cancel out

    since e also equals (a^-1)^-1 *a^-1) , then (a*a^(-1))= (a^-1)^-1 *a^(-1)


    ((a^(-1))^-1)=a for all a?

    Aren't suppose to multiply e on both sides as opposed to multiplying a?
    Last edited: Aug 24, 2007
  5. Aug 24, 2007 #4


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    Come on

    [tex]\left(a^{-1}\right)^{-1}\cdot\left(a^{-1}\cdot a \right)=e\cdot a [/tex]

    I used the associativity of the group operation. Now make the 2 multiplications, one inside the bracket and one in the RHS of the eq. What do u get ?
  6. Aug 24, 2007 #5

    matt grime

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    What is the definition of the inverse of a? It is certainly NOT 1/a.
  7. Aug 24, 2007 #6
    [tex]a^{-1}[/tex] is inverse of [tex]a[/tex], i.e. [tex]aa^{-1} = a^{-1} a = 1[/tex].
    So we see that [tex]a[/tex] is inverse of [tex]a^{-1}[/tex] which means, [tex](a^{-1})^{-1} = a[/tex].
  8. Aug 24, 2007 #7


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    no, it's no circular. He was explaining to you that you must use the definition of the (left) inverse. I.e., for any element [tex]g^{}[/tex] of the group, the inverse [tex]g^{-1}[/tex] satisfies
    g^{-1}\cdot g = e
    where the symbol [tex]e^{}[/tex] stands of the identity element of the group.

    Maybe the notation [tex]g^{-1}[/tex] for the inverse of [tex]g[/tex] is too familiar for its own good. You could try a different notation. For example, you could denote the inverse of [tex]g[/tex] as
    [tex]\bar g[/tex]. Then, the definition of the inverse says that
    \bar g \cdot g = e

    To prove the equality you mentioned, first simply rename the element [tex]g[/tex] to [tex]\bar a[/tex] (which is an element of the group if [tex]a[/tex] is, by definition) to see that
    \bar \bar a \cdot \bar a = e
    (which is just using the definition of inverse for [tex]\bar a[/tex]).

    Next, rename the element [tex]g[/tex] to [tex]a[/tex] to see that
    \bar a \cdot a = e
    (which, is just using the definition of inverse for a).

    If you multiply the above equation on the left by [tex]\bar \bar a[/tex] and the equation two-above on the right by [tex]a[/tex] you will see that the LH sides are equal and thus the RH sides are also equal which gives you the equality you desire.
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