Arjun Suresh (talk | contribs) |
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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: | + | 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 | + | (A) 2,2,3 |
| − | (B)3,3,3 | + | (B) 3,3,3 |
| − | (C)2,2,4 | + | (C) 2,2,4 |
| − | (D)2, | + | '''(D) 2,4,4''' |
| − | === | + | ==={{Template:Author|Arjun Suresh|{{arjunweb}} }}=== |
| + | [https://en.wikipedia.org/wiki/Abelian_group Abelian_group] | ||
| + | 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. | ||
| + | |||
| + | 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. | ||
| + | |||
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| + | 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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| + | {{Template:FBD}} | ||
| − | [[Category: | + | [[Category: Non-GATE Questions from Graph Theory]] |
| − | |||
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
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)
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
| e | e | a | b | c | |||||
|---|---|---|---|---|---|---|---|---|---|
| e | e | a | b | c | |||||
| a | a | e | b | b | b | b | c | e | e |
| c | c | b | e | a |
_____________________ |___|_e_|_a_|_b_|_c_| |_e_|_e_|_a_|_b_|_c_| |_a_|_a_|_e_|_b_|_b_| |_b_|_b_|_c_|_e_|_e_| |_c_|_c_|_b_|_e_|_a_|
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