0
2.0kviews
Show that if every element in a group is its own inverse, then the group must be abelian.
1 Answer
written 7.9 years ago by |
Let (G,o) is a group.
$\therefore \ \ if \ \ a, b \in G \ \ then \ \ a^{-1}, b^{-1} \in G$
also if $aob \in G$ the $(aob)^{-1} \in G$
But we have $a=a^{-1}$ and $b=b^{-1}$
As such $(aob)=(aob)^{-1}=b^{-1}oa^{-1}=(boa)$
i.e, (G,o) is commutative. Hence (G,o) is abelian.