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Math Help - prove that (a_1a_2...a_n)^2=e for every finite abelian group.

  1. #1
    No one in Particular VonNemo19's Avatar
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    prove that (a_1a_2...a_n)^2=e for every finite abelian group.

    OK. I'm trying, but I keep running into problems. I've recently proven that

    ab=e\Rightarrow{ba}=e and a=a^{-1}\Leftrightarrow{a}^2=e. What I'm asked to prove seems to be a generalization of of these two statements, but I can't quite see it. Is there any way someone could point me in the right direction here?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: prove that (a_1a_2...a_n)^2=e for every finite abelian group.

    Quote Originally Posted by VonNemo19 View Post
    OK. I'm trying, but I keep running into problems. I've recently proven that

    ab=e\Rightarrow{ba}=e and a=a^{-1}\Leftrightarrow{a}^2=e. What I'm asked to prove seems to be a generalization of of these two statements, but I can't quite see it. Is there any way someone could point me in the right direction here?
    This isn't true 1+0=1 in \mathbb{Z}_2. The correct statement is that the product over all the elements in an abelian group of odd order is the identity. Hint: pair things with their inverse (how do you know the inverse isn't equal to the original element)?


    EDIT: Misread problem, my apologies.
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  3. #3
    MHF Contributor

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    Re: prove that (a_1a_2...a_n)^2=e for every finite abelian group.

    what you are actually being asked to prove is that, if you multiply every group element (in an abelian group, the abelian-ness is important, here) together, and square the result, you get the identity.

    there are, actually, LOTS of ways to prove this.

    here is one approach:

    prove that (a_1a_2\dots a_n)^{-1} = (a_1)^{-1}(a_2)^{-1}\dots(a_n)^{-1} (this uses the fact that G is abelian).

    then show that (a_1)^{-1}(a_2)^{-1}\dots(a_n)^{-1} = a_1a_2\dots a_n (why? show that map a_i \to (a_i)^{-1} is bijective)

    and now....you're done (why?).

    EDIT: alex, you missed the sqaured.
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