Question

If every element of a group is its own inverse then prove that the group is abelian

Answer 1

Let $G$ be a group and $a,b\in G$.
Since every element of a group is its own inverse,
We have $a=a^{-1}$ and $b=b^{-1}$ .....$\left(1\right)$
Also, $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}[\because (ab{)}^{-1}=b^{-1}{a}^{-1}]$
$=b.a$ [using $\left(1\right)$]
$ab=ba$ for all $a,b\in G$
Hence, $G$ in an abelian group.