Answer to Question #13177 in Abstract Algebra for Melvin Henriksen

Question

Prove that if all squares of all elements are equal identity of group, that it is abelian.

Answer 1

Notice that for any two elements a,b in a group G their commutant
[a,b]
= a b a^{-1} b^{-1}
is generated by squares:

[a,b] = a b a^{-1}
b^{-1} =
ab ab b^{-1} a^{-1} a^{-1} b^{-1} =

(ab)^2 b^{-1} a^{-1} b b^{-1} a^{-1} b b^{-1} b^{-1} =

(ab)^2 ( b^{-1} a^{-1} b ) ( b^{-1} a^{-1} b ) ( b^{-1} b^{-1})
=
(ab)^2 (b^{-1} a^{-1} b)^2 b^{-2}

Hence if all squares of
all elements are equal identity of group G,
then all commutants in G are
identity, i.e. all elements in G commute, that is G is abelian.