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In a finite group G, the inverse of each element is a power of itself.

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    If G is a finite group and g is in G, then there exists a positive integer r with g^r=e.
    and in general, prove that, in any finite group G, the inverse of each element is a power of itself.


    2. Relevant equations



    3. The attempt at a solution
    I know if a group is finite then it has a finite order, which means g^n=e, but how do you write that as a proof?
     
  2. jcsd
  3. Oct 11, 2009 #2

    CompuChip

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    Be careful: that the order of the group is finite does not automatically mean that g^n = e for some n.

    Did you learn about the Lagrange theorem yet?
    It says that the order of any element of the group, divides the order of the group. When you apply that you know that in particular, g^n = e for some finite number n and you can write down an explicit inverse for g (which, as you probably know, is unique).
     
  4. Oct 11, 2009 #3
    What about:

    Assume on the contrary that there does not exist any [tex]k \in \mathbb{Z}^{+}[/tex] such that, for [tex]g \in G[/tex], [tex]g^{k} = 1_{G}[/tex], the identity. As G is finite, we can't have all [tex]g^{k}[/tex] distinct for all positive integers k, as it would result in a set having infinite cardinality. Thus [tex] \exists k_{1}, k_{2} \in \mathbb{Z}_{+}[/tex] such that [tex]k_{1} \neq k_{2}, g^{k_{1}} = g^{k_{2}}[/tex]. Without loss of generality we may assume that [tex] k_{1} > k_{2}[/tex]. But then [tex]g^{k_{1}}g^{-k_{2}} = g^{k_{2}}g^{-k{2}} \Leftrightarrow g^{k_{1}-k_{2}} = 1_{G}[/tex], and [tex]g_{1}-g_{2} \in \mathbb{Z}^{+}[/tex], which is a contradiction. Thus the assumtion that such an integer k exists must be false, and the assertion is proven.

    This is actually pretty much what you stated, but formulated in such a way that it provides a proof. Though, one must be careful. For example, this proof uses the "laws of exponents" for groups, which has to be proven. Though, I suspect you have already done that in your course.
     
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