# Answer to Question #13177 in Abstract Algebra for Melvin Henriksen

Question #13177

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

Expert's answer

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.

[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.

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