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Show that if every element in a group is its own inverse, then the group must be abelian.

written 8.2 years ago by teamques10 ★ 69k

modified 3.1 years ago by sagarkolekar ★ 11k

discrete mathematics

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written 8.2 years ago by teamques10 ★ 69k

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.

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