# Let $(G,\circ)$ be a group in which $x\circ x$ is the neutral element for all $x \in G$ [duplicate]

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
@awllower oh yeah I must have mistagged it –  Adam Feb 13 '13 at 13:18
Haha. A fun mistake indeed. –  awllower Feb 13 '13 at 13:19
Isn't this a duplicate? –  1015 Feb 13 '13 at 13:39

## marked as duplicate by Derek Holt, Henry T. Horton, Amzoti, Henning Makholm, Cameron BuieFeb 13 '13 at 15:58

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

$$xy=x(xy)^2y=xxyxyy=(xx)yx(yy)=yx.$$

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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
@Adam Since $x=x^{-1}$ for all $x\in G$ and $xy$ is also element of $G$, so $xy=(xy)^{-1}$. –  tetori Feb 13 '13 at 14:34

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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Let $x,y \in G$. Since $x^{-1}=x$ and $y^{-1}=y$, $[x,y]=xyx^{-1}y^{-1}=xyxy=(xy)^2=1$.
What is $(x\circ y)^2$? And what does that mean?