Proving that a group $(G, \ast)$ is abelian if $x^3=x$ for all $x\in G$

Question

If $(G, \ast)$ is a group so that $x^3=x$ for all $x\in G$ then $G$ is abelian

Answer 1

Note that the assumption is equivalent to $x^2=e$ for all $x\in G$. Let $x,y\in G$. Then

$$xy=y^2 x y x^2=y(yx)(yx)x=yx.$$

Answer 2

We have $$x^3 = x \Longrightarrow x^2 = 1 \Longrightarrow x = x^{-1}$$

for all $x$. Thus $$ab = (ab)^{-1} = b^{-1}a^{-1} = ba$$ for all $b , a \in G $.

Answer 3

$$x^2 = e ,\forall x \in X \Rightarrow (xy)^2 = x^2y^2 \Rightarrow (xy)(xy) = x^2y^2 \Rightarrow xyx = x^2y \Rightarrow yx=xy.$$

For $x,y \in G$ of course.

Answer 4

Hints:

1) if $x^3 = x$, then $(xy)^3 = xy$. Remember that you can't say yet that $(xy)^3 = x^3 y^3$.

2) Apply $x^{-1}$ to the given relation to get $x^2 = e$. Now do it again. What is the relation between $x$ and $x^{-1}$?