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Thread: Abelian groups

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    Abelian groups

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

    So what I need to show is that, in addition to G satisfying closure, associativity, identity and inverse, $\displaystyle G$ also satisfies commutativity ($\displaystyle ab=ba, \forall a,b \in G)$ 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" $\displaystyle [x,y]=xyx^{-1}y^{-1}$ of two elements. We have $\displaystyle [x,y]=1$ if and only if $\displaystyle x,y$ commute.

    In this case, note that we have $\displaystyle x^2=1 \ \forall x \in G$. Thus we have $\displaystyle [x,y]=xyx^{-1}y^{-1}=xyxy=(xy)^2=1$ 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 $\displaystyle g^{-1}=g ~\forall g \in G$. Prove that $\displaystyle G$ is abelian.

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