Abelian Group Revision as of 22:45, 9 December 2013 by Arjun Suresh (talk | contribs) (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

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