# How to prove that if each element of group is inverse to itself then group commutative? [duplicate]

How can you prove that if each element of group is inverse to itself then the group is commutative?

• 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