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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
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4 Answers

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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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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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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