Prove that $x=x^{-1}$ for all $x \in G$ and that G is commutative.
I have no idea where to start. I know that commutative will mean that for
$x,y \in G$
$x\circ y = y \circ x$
but I don't know how to prove that.
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I think the tag should be groups, instead of algebraic groups, for it is a totally different topic, IIRC. – awllower Feb 13 '13 at 13:09
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@awllower oh yeah I must have mistagged it – Adam Feb 13 '13 at 13:18
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Haha. A fun mistake indeed. – awllower Feb 13 '13 at 13:19
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Isn't this a duplicate? – Julien Feb 13 '13 at 13:39
4 Answers
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Since any element of $G$ satisfies $x^2=e$ ($e$ is neutral element), so $xy$ also satisfies $(xy)^2=e$, if $x$, $y\in G$. And
$$xy=x(xy)^2y=xxyxyy=(xx)yx(yy)=yx.$$
Hanul Jeon
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How would I show that $x=x^{-1}$ would it have something to do with $(x \circ y)^{-1}$ – Adam Feb 13 '13 at 14:03
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@Adam Since $x=x^{-1}$ for all $x\in G$ and $xy$ is also element of $G$, so $xy=(xy)^{-1}$. – Hanul Jeon Feb 13 '13 at 14:34
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For any $x\in G$, $x^2=e$, so $x=x(xx^{-1})=(xx)x^{-1}=x^{-1}$.
For any $x,y\in G$, $xy=x^{-1}y^{-1}=(yx)^{-1}=yx$, where the first and last equalities follow from the previous paragraph.
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Hint:
What is $(x\circ y)^2$? And what does that mean?
Stefan Hansen
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awllower
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Let $x,y \in G$. Since $x^{-1}=x$ and $y^{-1}=y$, $[x,y]=xyx^{-1}y^{-1}=xyxy=(xy)^2=1$.
Seirios
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But that is my answer, in hint form. In any case, thanks for the answer.Sorry if this claim is too rude: I will delete it if needed. – awllower Feb 13 '13 at 13:20