A certain
group is known to have the property that every element is its own inverse.
Prove that the group is commutative.
I studied groups for two whole weeks in 6th grade. Does this work?
A member of the group can only have one inverse in the group. (Rule I)
Call the identity element "i", and call the operation "!". Then:
(A ! B) ?= (B ! A) (for all A and B)
(A ! B) ! (A ! B) ?= (A ! B) ! (B ! A)
i = (A ! B) ! (B ! A)
From the fact that every element is its own inverse and Rule I:
(A ! B) = (B ! A)
QED
Edited on February 26, 2023, 10:24 am
possible solution
