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Suppose that we have some group G in which the following holds for all $x $ belongs to G: $x^{-1} =x$. Prove that G is abelian.

How would you guys solve this one?

lulu
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TWstud
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    Hint: Just write down the commutator of a general pair $a,b\in G$. – lulu Feb 18 '20 at 20:52
  • $xy=(xy)^{-1}=y^{-1}x^{-1}=yx$. The first equation is the assumption applied to $xy$, while the last is the assumption applied to $y$ and to $x$. –  Feb 18 '20 at 20:55
  • @lepidon: I honestly didn't see your comment before I posted, as an answer, exactly what you wrote – J. W. Tanner Feb 18 '20 at 20:57
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    @J.W.Tanner Even if you did, that is not a problem. This is not a new discovery or anything like that. It is just an exercise. –  Feb 18 '20 at 20:59
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    @J.W.Tanner By the way, you didn't get downvoted by me, but by who commented in your answer. I was the one who up-voted it and all other answers. –  Feb 18 '20 at 21:00

3 Answers3

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Well, we have $$xy = (xy)^{-1} = y^{-1}x^{-1} = yx$$ for all $x,y \in G$. Hence $G$ is abelian.

Con
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Abelian or not: $(ab)^{-1} = b^{-1}a^{-1}$.

(Because 1) $(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = a*e*a^{-1}= aa^{-1} =e$ and 2) inverses are unique (because $fa = af =e$ and $ga=ag = e$ means $fag = fe =f$ and $fag =eg=g$ so $f=g$))

But we are told $x^{-1} = x$ so $(ab)^{-1}=ab$ and $b^{-1} = b$ and $a^{-1} = a$ so....

J. W. Tanner
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fleablood
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Take $x,y \in G$ , you want to show that $xy = yx$.

Or , the commutator of $x,y$ is the identity , that is $[x,y] = xyx^{-1}y^{-1}=e$.

Can you take this from here?

infinity
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