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Prove an element and its inverse have same order

  1. Feb 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that in any group, an element and its inverse have the same order.

    2. Relevant equations

    3. The attempt at a solution
    I know that I need to show |g|=|g^-1|. This is how I approached it but I'm not sure I'm right, it seems like I'm missing something.
    Let |g|=m and |g^-1|=n for g, g^-1 elements of G. Then (g)^m=e and (g^-1)^n=e. Therefore, either (g)^m <= (g^-1)^n or (g)^m >= (g^-1)^n. If (g)^m <= (g^-1)^n then e<=e and m<=n. If (g)^m >= (g^-1)^n then e>=e and m >= n. Therefore m = n and |g|=|g^-1|.
    Last edited: Feb 14, 2011
  2. jcsd
  3. Feb 14, 2011 #2


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    It's right in spirit. But lacks in presentation. Sure, if x^k=e then (x^(-1))^k=e^(-1)=e. Now just use the definition of '|x|'. Don't use '<' for the elements of the group. Groups generally aren't ordered.
    Last edited: Feb 14, 2011
  4. Feb 14, 2011 #3

    I might be misreading your attempted solution, but I don't understand what you mean by f^m <= (g^-1)^n.

    Here is a start: Let m = ord(g) and n = ord(g^-1). Then g^m = (g^-1)^n. Now, multiply on the left by g^n to get: (g^n)(g^m) = e. Since g^m=e, we must have that g^n = e, from which we see that m|n (that means m divides n.) Now, all you need to do is to show that n|m (n divides m) to show that m=n. To do that, you have to do something really similar to what I did.
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