edited by
1,723 views
14 votes
14 votes
Show that if $G$ is a group such that $(a. b)^2 = a^2.b^2$ for all $a, b$ belonging to $G$, then $G$ is an abelian.
edited by

2 Answers

Best answer
20 votes
20 votes
$\text{If}$ $ (a*b)^{2}=a^{2}*b^{2}$ $ \forall a,b \in G$   $\text{(*$ $ is an operator})$

$\text{LHS}\Rightarrow (a*b)*(a*b)$

$\qquad = a*b*a*b$

$\text{RHS}\Rightarrow a*a*b*b$

$\text{Comparing LHS and RHS}$

$\implies a*b*a*b=a*a*b*b$

$\text{Applying Cancellation law}$

 $b*a=a*b \text{ (Commutative law)}$

$\text{So, it is Abelian group.}$  $\text{(It is a group and it follows Commutative property)}$
edited by
2 votes
2 votes

G is a group

so, it will follow cancellation and associative rule

    (a . b)² = a². b²

⇒ a . (b . a) .b = a . a . b . b [using associative rule]

a . (b . a) .b = a . a . b . b [using cancelation rule]

⇒ b . a = a . b [commutative]

As it is already a group and now its following commutative property it is abelian group

Related questions

1 votes
1 votes
1 answer
1
go_editor asked Dec 20, 2016
537 views
Verify whether the following mapping is a homomorphism. If so, determine its kernel.$f(x)=x^3$, for all $x$ belonging to $G$.
1 votes
1 votes
0 answers
2
go_editor asked Dec 20, 2016
441 views
Verify whether the following mapping is a homomorphism. If so, determine its kernel.$G$ is the group of non zero real numbers under multiplication.
1 votes
1 votes
0 answers
3
go_editor asked Dec 20, 2016
477 views
Verify whether the following mapping is a homomorphism. If so, determine its kernel.$\overline{G}=G$
25 votes
25 votes
2 answers
4
go_editor asked Dec 20, 2016
2,554 views
If the set $S$ has a finite number of elements, prove that if $f$ maps $S$ onto $S$, then $f$ is one-to-one.