How to prove that if each element of group is inverse to itself then group commutative? How can you prove that if each element of group is inverse to itself then the group is commutative?
 A: HINT: If $a,b\in G$, then $(ab)^2=1_G=a^2b^2$.
A: Let $g \in G$; we know that $g = g^{-1}$, and that $g^{-1} \in G$ because groups contain the inverse of each of their elements. Now suppose $x,y \in G$. Then $xy \in G$ by closure under products, and so $xy = (xy)^{-1} = y^{-1}x^{-1} = yx$, where we used the fact the $x=x^{-1}$ and $y = y^{-1}$ in the last step. This shows that $G$ is a commutative group.
A: The group being commutative means that $ \forall a, b \in G$, $ab = ba$. Since G is a group, $ab \in G$ and so is it's own inverse, which means $(ab)(ab) = 1$ multiplying on the left by a and then by b gives $bab = a$ and then $ab = ba$ as required.
A: Since you've got it, $$xy = \overline {xy} = \overline y \overline x = yx$$
A: Since the problem is quite elementary, I'll give you a hint:
being one's own inverse means that $a^2=1$ for all $a$, what do you know about the identity of a group? 
A: say a and b belong to G
a=(a)^-1   and 
b=(b)^-1
=> (axb) = ((a^-1) x (b^-1))
=>  axb = (bxa)^-1        -------------------------------(1)
let c=bxa
it is obvious from closure prop. that c belongs to G
               so c=(c)^-1
               or (bxa)=(bxa)^-1      ---------------------(2)
from (1) and (2)   axb=bxa
Hence it is a commutative group
