Let <math>G\{e,a,b,c\}</math> be an abelian group with <math>'e'</math> as an identity element. The order of the other elements are:

(A) 2,2,3

(B) 3,3,3

(C) 2,2,4

(D) 2,4,4

Solution by Arjun Suresh

Abelian_group

As a consequence of Lagrange's theorem, the order of every element of a group divides the order of the group. Hence, 3 cannot be the order of any element in the group.

Now, consider 2,2,4. If it has to hold then,

a * a = e

b * b = e and

c * c = a

=> a * c = b and

b * c = e (to get $c^4 = e$)

But then, the associativity property of

[(a * c) * b] = [a * (c * b)] 

fails as (a * c) * b = e and a * (c * b) = a. Hence, 2,2,4 is not the answer.

* e a b c
e e a b c
a a e c b
b b c a e
c c b e a
a * a = e => order(a) = 2
b * b * b * b = a * b * b = c * b = e => order(b) = 4
c * c * c * c = a * c * c = b * c = e => order(c) = 4

So, the answer is 2,4,4. (2,2,2 is another possibility)




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