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How do we find whether the group is finite or infinte in questions like below ?

If (G,*) is a group such that $(a * b)^{2} = (a*a) *(b*b)$ for all a,b belonging to G, then G is

a) Finite group

b) Infinite group

c) Abelian group

d) None of the above
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Here is the definition.

Let a be an element of the group G. If there exists a positive integer n such that an = e, then a is said to have Finite Order, and the smallest such positive integer is called the order of a, denoted by o(a)If there does not exist a positive integer n such that an = e, then a is said to have Infinite Order.

There is no way to prove that the given group (G,*) is a finite group or not. Because It does not talk about the sets. I have read this document. You can also read if you want to see that.

Now In the above question, The correct answer is C. The given group is an Abelien group. A group with Commutative property is called Abelian group. Because we will get (a * b)^2 = (a * a) * (b * b), only when the group is commutative. Like this:

(a * b)^2 = (a * b) * (a * b)
= (a * b) * (b * a)    // Commutative property

= a * ((b * b) * a)    // Associative property

= a * (a * (b * b))    // Commutative property

= (a * a) * (b * b)    // Associative Property

Hence the given group is Abelian group. Hence Option C is the correct answer.

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@rude but here u havenot proved identity and inverse property

@Srestha it is given that (G,*) is a group, it means that it follows Closure, Associativity, Identity and inverse. I have to prove only Commutative property. Which I have shows that to achieve above expression you must have commutative property true else it will not be possible.

yes

A group with finite number of element is called as finite group and it represented by O(G).

let (G,*) be a group and a belong to G,Then order of an element 'a' is the smallest +ve integer n such that a^n=e is identitiy element;

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

Applying right & left cancellation law

b*a=a*b

that's why above group is abelian group.