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the number of generators of the group { 0,1,2........... 14} under the group operation addition modulu 15 is
asked in Set Theory & Algebra by Veteran (13k points)   | 160 views
any formula for this??
1 and 2 are definitly a generator

2 Answers

+7 votes
Best answer

$\left ( Z,+_m \right )$ is a standard cyclic group.

And no of generators in a cyclic group is = $\phi(n)$ where n is the order of the group.

$\phi(15) = \phi(3^1 * 5^1) = (3^1-3^0)*(5^1-5^0) = 2* 4 = 8$ generators. 


PS: how to calculate $\phi\left ( n \right ) = \phi\left ( p_1^{x_1}.p_2^{x_2} \right ) = \ \left ( p_1^{x_1} -p_1^{x_1-1} \right ).\left ( p_2^{x_2} -p_2^{x_2-1} \right )$

 

answered by Veteran (47k points)  
selected by
how did u choose 3 and 5 here??
prime factorization
+3 votes

The number of generators in a group is based on the order of the group.

The no. of generators of a group will always be less than the order of the group and co-prime to the order of the group.

Here O(g) = 15,

Co-primes of 15 are 1,2,4,7,8,11,13,14

Thus the number of generators for this group is 8.

answered by Active (1.2k points)  
@nitish it will work for addition modulo or for other operation ?
@wanted It will work if the group is a cyclic group.
can u tell in brief about cyclic group ...i know they are abelian group or much ?
G = $\left \{ a^{i} | 1\leq i\leq n \right \}$

G is a group in which all the elements of the group can be written as power of one element. If this property holds then G is a cyclic group.

'a' is the generator of the group.

Ex {1,2,3,4,5,6} mod 7 is cyclic group where 3 and 5 are generators.


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