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?

-
This has to be a duplicate! I have searched though, but to no avail. The closest I could find was this question. – user1729 Oct 26 '12 at 8:58
Plenty of answers, but I'm the only one who's up-voted the question so far. – Michael Hardy Jun 9 '14 at 19:02

HINT: If $a,b\in G$, then $(ab)^2=1_G=a^2b^2$.

-

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.

-
IMHO, the most elegant answer. – beroal Mar 24 '15 at 12:08

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.

-

Since you've got it, $$xy = \overline {xy} = \overline y \overline x = yx$$

-
Please consider using $\LaTeX$ to format your answers. – Ali Caglayan Oct 22 '14 at 8:59
@Alizter Thank you for the reference: I didn't know how. – Thumbnail Oct 22 '14 at 9:20

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?

-

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

-