Let $A$ be a group, where $a^2=1$, a belongs to $A$. Prove that this group is commutative. [duplicate]
Let $a,b$ be elements of the group. We want to show that $a * b = b * a$. But this is equivalent to $a * b * a = b$ which is equivalent to $(a * b) * (a * b) = 1$, which is true by our hypothesis that $x * x = 1 $ for every $x$ in the group.
We have: $x^2 = e$, and $y^2 = e$. So: $(xy)^2 = e = e*e = x^2y^2$. So $(xy)^2 = x^2y^2$. Thus: $xyxy = xxyy$. So: $x^{-1}xyxyy^{-1} = x^{-1}xxyyy^{-1}$. Since $x^{-1}x = yy^{-1} = e$, and $ex = x$, $ey = y$, you have : $yx = xy$ for all $x, y$ proving the group commutative.