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  1. MHB Apprentice

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    #1
    Here is a problem from some russian book of algebra:
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    Suppose $ \displaystyle G$ is a finite group. An automorphism $ \displaystyle \varphi$ "operates" on this group. This automorphism satisfies the following two conditions: 1) $ \displaystyle \varphi^2=e_G$; 2) if $ \displaystyle a\not= e$, then $ \displaystyle \varphi(a)\not= a.$ Prove that $ \displaystyle G$ is an abelian odd group.
    $ \displaystyle \varphi(x)=y\leftrightarrow\varphi(y)=x$ and I know $ \displaystyle \varphi(e)=e.$ I can see from this that $ \displaystyle G$ is a group of odd order. How I prove commutativity? Do you think I can prove first that $ \displaystyle \varphi(a)=a^{-1}$?

  2. MHB Craftsman
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    #2
    Hi,
    This is a "standard" exercise. Here is a link to a proof:

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    #3
    Quote Originally Posted by Andrei View Post
    Here is a problem from some russian book of algebra:


    $ \displaystyle \varphi(x)=y\leftrightarrow\varphi(y)=x$ and I know $ \displaystyle \varphi(e)=e.$ I can see from this that $ \displaystyle G$ is a group of odd order. How I prove commutativity? Do you think I can prove first that $ \displaystyle \varphi(a)=a^{-1}$?
    as johng's post shows, the answer is yes.

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