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Que: Which one of the following is not true?

a) In a group (G,*), if a*a = a, then a=e, where e is an identity element.

b) In a group (G,*), if x-1 = x,  ∀x ∈ G, then G is an Abelian group.

c) In a group (G,*), if (a*b)2 = (a2*b2) then G is an Abelian group.

d)  In a group (G,*), if (a*b)n = (an*bn) then G is an Abelian group, where n={2, 3, 4, ......}

PS: c) and d) seem correct to me.

2 Answers

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As you know c and d are correct let us examine a & b.

b)  let a = a-1

 (ab)(ab)-1 = e

(ab)(ab) = e                // a = a-1

(ab)(a-1b-1) = e

(aa-1)(bb-1) = e

ee = e

e = e

True..

Hence a) is going to be false.

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${\color{Red} All}$ are ${\color{Magenta}True }$

a) In a group (G,*), if a*a = a, then a=e, where e is an identity element.


We will prove this by contradiction, $(a*a = a) \rightarrow a = e$

let us assume that $a \neq e$ is true 
that means a is not the identity element

According to our assumption there must exist some identity element (e) that satisfies
a*e1 = a or e2 * a = a  

e1 = e2 = e ( we know that * is commutative and associative) 
and we assumed that $a \neq  e$ but if $(a*a = a)$ is true then $a \neq  e$ can't be true because 
$(a*a = a)$ a acts as both e1 and e2

b) In a group (G,*), if x-1 = x,  ∀x ∈ G, then G is an Abelian group.

An abelian group is a group in which the law of composition is commutative.

option b says that every element if set G is inverse of itself. This is TRUE 
Assume G = {−1,1} 

c) In a group (G,*), if (a*b)2 = (a2*b2) then G is an Abelian group.

assume G be any set of integers 
Let G = {1,2},
(1*2) = (1* 22

d)  In a group (G,*), if (a*b)n = (an*bn) then G is an Abelian group, where n={2, 3, 4, ......}

Same reasoning as option c


 

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