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

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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 at 19:02

6 Answers 6

up vote 3 down vote accepted

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

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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.

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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.

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

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

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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

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