How do you prove that a group #G# is abelian if and only if #a^2 b^2 = (ab)^2# for all #a, b in G# ?

2 Answers
Aritra G.
Sep 2, 2017

If a group #G# is Abelian, then for any two elements #a# and #b#,

#a*b = b*a#

Now,

Define #c=a*b# which is also a member of #G# on virtue of closure property.

Then #c^2 = c*c#
#implies (a*b)^2=(a*b)*(a*b)#

By associative axiom,

#(a*b)^2 = a*(b*a)*b#

But, by property of commutativity (for an Abelian group),

#implies (a*b)^2=a*(a*b)*b#
#implies (a*b)^2=(a*a)*(b*b)#

Thus we get our final result,

#(a*b)^2 = a^2*b^2#

For two arbitrary elements #a# and #b# of an Abelian group #G#.

George C.
Sep 3, 2017

See explanation:

Explanation:

Since group multiplication is associative I will omit the allowed rearrangements of parentheses that indicate performing multiplications in different orders.

Suppose #G# is abelian.

Then for all #a, b in G#:

#ab=ba#

Then, using (associativity and) commutativity we find:

#(ab)^2 = acolor(blue)(ba)b = acolor(blue)(ab)b = a^2b^2#

Conversely, suppose that for all #a, b in G#:

#(ab)^2 = a^2b^2#

Then for all #a, b in G#:

#ba = a^(-1)color(blue)(abab)b^(-1) = a^(-1)color(blue)(a^2b^2)b^(-1) = ab#

i.e. #G# is abelian.

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