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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:
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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  
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(A) 2,2,3  
  
(B)3,3,3  
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(B) 3,3,3  
  
(C)2,2,4  
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(C) 2,2,4  
  
(D)2,3,4
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'''(D) 2,4,4'''
  
===Solution===
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==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
_____________________
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[https://en.wikipedia.org/wiki/Abelian_group Abelian_group]
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|_e_|_e_|_a_|_b_|_c_|
 
|_a_|_a_|_e_|_b_|_b_|
 
|_b_|_b_|_c_|_e_|_e_|
 
|_c_|_c_|_b_|_e_|_a_|
 
  
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As a consequence of [https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory) 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.
  
[[Category:Other Questions]]
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Now, consider 2,2,4. If it has to hold then,
[[Category: Algorithms & Data Structures]]
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a * a = e
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b * b = e and
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c * c = a
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=> a * c = b and
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b * c = e (to get $c^4 = e$)
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But then, the associativity property of
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[(a * c) * b] = [a * (c * b)]
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fails as (a * c) * b = e and a * (c * b) = a. Hence, 2,2,4 is not the answer.
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{| class="wikitable" style="text-align: center;background-color: #ffffff;" width="35%"
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!*
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!e
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!a
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!b
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!c
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|-
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!e
 +
|e
 +
|a
 +
|b
 +
|c
 +
|-
 +
!a
 +
|a
 +
|e
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|c
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|b
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|-
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!b
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|b
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|c
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|a
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|e
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|-
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!c
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|c
 +
|b
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|e
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|a
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|}
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a * a = e => order(a) = 2
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b * b * b * b = a * b * b = c * b = e => order(b) = 4
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c * c * c * c = a * c * c = b * c = e => order(c) = 4
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So, the answer is 2,4,4. (2,2,2 is another possibility)
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{{Template:FBD}}
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[[Category: Non-GATE Questions from Graph Theory]]

Latest revision as of 11:37, 15 July 2014

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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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)2,3,4

Solution[edit]

_____________________ |___|_e_|_a_|_b_|_c_| |_e_|_e_|_a_|_b_|_c_| |_a_|_a_|_e_|_b_|_b_| |_b_|_b_|_c_|_e_|_e_| |_c_|_c_|_b_|_e_|_a_|