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Math Help - Abelian groups

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    Abelian groups

    Suppose that G is a group in which g^{-1}=g ~\forall g \in G. Prove that G is abelian.

    So what I need to show is that, in addition to G satisfying closure, associativity, identity and inverse, G also satisfies commutativity ( ab=ba, \forall a,b \in G) but I'm not sure how to get started.
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    MHF Contributor Bruno J.'s Avatar
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    In a general group, consider the "commutator" [x,y]=xyx^{-1}y^{-1} of two elements. We have [x,y]=1 if and only if x,y commute.

    In this case, note that we have x^2=1 \ \forall x \in G. Thus we have [x,y]=xyx^{-1}y^{-1}=xyxy=(xy)^2=1 and we are done.
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    That's quite a nice way of doing this actually. We haven't studied commutators yet but it makes sense, thanks!
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by nmatthies1 View Post
    That's quite a nice way of doing this actually. We haven't studied commutators yet but it makes sense, thanks!
    You're welcome! Good luck.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nmatthies1 View Post
    Suppose that G is a group in which g^{-1}=g ~\forall g \in G. Prove that G is abelian.

    So what I need to show is that, in addition to G satisfying closure, associativity, identity and inverse, G also satisfies commutativity ( ab=ba, \forall a,b \in G) but I'm not sure how to get started.
    Merely note that for any arbitrary g,g'\in G we have that gg'\in G, and by assumption \left(gg'\right)^2=e. So, gg'=g\left(gg'\right)^2g'=\left(gg\right)g'g\left(  g'g'\right)=g^2g'gg'^2=g'g
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