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IF ( G, * ) is an abelian group then discuss correctness of each of the options ...

a)  a = a-1 for all a $\epsilon$ G

b) a2 = a for all a $\epsilon$ G

c) (a*b)2 = a2 * b2 for all a,b $\epsilon$ G

d) G is finite order

My Approach

Abelian Group -- > Commutative Group

A) 

Let  (a*b)-1= b-1 * a-1

    givent that a-1 = a

   Therefore , (a*b)-1 = b * a

  Therefore , G is a abelian group   Given statement is correct

B)

a2 = a 
    IF a*a = a THEN a is identity element (e)

 

C)

( a*b)= a2 * b2

 (a*b)*(a*b) = (a*a) * (b*b)

a* (b*a)*b = a*(a*b)*b

  b*a = a*b   // cancelling a using left cancellation law and b using right cancellation law

  Therefore G is abelian group Given statement is correct

D)


PLEASE HELP ME ..

asked in Set Theory & Algebra by Veteran (22.8k points) 46 216 358 | 209 views
what are you proving here?
@arjun sir , chechking the correctness of statements given
But you are asked to prove statement 1. Now, you assumed statement 1 true for A and took another assumption and proved G is abelian. It should be done like this: G is abelian, use properties of abelian group and show that statement A is true. Or just give a counter example showing statement A is false.
Ok. sir

2 Answers

0 votes
D. We have to prove that every |G| is finite for abelian groups.

Since all abelian groups are cyclic i.e a^n = a; for some n.

now consider n is infinite, which implies this is not a cyclic group, hence not an abelian. Therefore n is finite.

Thus |G| = n which is finite.
answered by Active (1.6k points) 3 13 31
–1 vote
Youb r right abolutely,then why r u asking ?
answered by Boss (5.6k points) 3 9 32


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