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Is the following statement true or false:

If $G$ is a group with the property that $g=g^{-1}$ for all $g \in G$, then $G$ is abelian.

I believe it is false since I know that abelian or commutative property implies that every element in $G$ must have an inverse. Thus $g\cdot g^{-1}= g^{1}\cdot g = e$.

I need someone to check my attempt.

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Lynnie
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  • I did not understand i know that abelian or commutative property implies that every element in G must have an inverse. Thus g.g^(-1)= g^(1).g = e –  Sep 09 '14 at 09:23
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    Hint: If $G$ is cyclic, then $G$ is abelian. Now, what do you know about cyclic groups? – Jose Arnaldo Bebita Dris Sep 09 '14 at 09:25
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    Not clear how your attempt is meant to work. You need to prove that the group (you know it is a group, so you know inverses exist) is abelian. Or else, you need to come up with an example where $g=g^{-1}$ for all $g$ and the group is not abelian. Which one did you attempt? – Jeppe Stig Nielsen Sep 09 '14 at 09:28
  • @PraphullaKoushik That is my attempt but i want comments on the attempt and if the attempt i need know if am on the right track or not. – Lynnie Sep 09 '14 at 09:30
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    your attempted proof is not a proof at all. First, the statement is true. Second, that there exists an nverse (which in fact is unique) for any element in the group is one of the axioms of group theory and thus true for any group, abelian or not. – Timbuc Sep 09 '14 at 09:37
  • @Lynnie : I am saying that i did not understand your attempt... –  Sep 09 '14 at 09:38
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    See also Group where every element is order 2 and the threads linked to it. – Jeppe Stig Nielsen Sep 09 '14 at 09:44
  • @Timbuc thanks thats clear enough. – Lynnie Sep 09 '14 at 09:50

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Your attempt is not valid. You have stated the axioms of inverses for groups, but these have nothing to do with whether the group is abelian or not. The question is when every element is its own inverse, which happens only rarely.

If $G$ is a group such that $g^{-1}=g$ for all $g\in G$, how can we show that any two elements of $G$ commute? If $x,y\in G$, what happens if we set $g=xy$?

Andrew Dudzik
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Hints for you to understand and prove:

== In any group, $\;(gh)^{-1}=h^{-1}g^{-1}\;$ :

== A group is abelian iff $\;\forall\;g,h\in G\;,\;\;gh=hg\iff\;(hg)^{-1}gh=1\;$

Timbuc
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Suppose that $g=g^{-1}, \forall g \in G$. Then

$g.g=g.g^{-1}=e\implies g^2=e.$

So for $x,y\in G, (xy)(xy)=e$.

$\therefore xy=(xe)y=x(xy.xy)y=(xx)yx(yy)=yx$.

Hence G is abelian. $\blacksquare$

creative
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  • Hi can you explain this g.g=g.g^(−1)=e⟹g^(2)=e. – Lynnie Sep 09 '14 at 10:05
  • Axiom 4 : $\exists$ unique $a^{-1}\in G$ such that $a.a^{-1}=a^{-1}.a=e$. Here considering step 1 and multiply both sides by g gives LHS as $g^2$ and RHS as $e$. – creative Sep 09 '14 at 10:13