Question

the number of generators of the group { 0,1,2........... 14} under the group operation addition modulu 15 is

Comments

any formula for this??
1 and 2 are definitly a generator

Answer 1

$\left ( Z,+_m \right )$ 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. 

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

Comments

how did u choose 3 and 5 here??

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prime factorization

Answer 2

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.

Comments

@nitish it will work for addition modulo or for other operation ?

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@wanted It will work if the group is a cyclic group.

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can u tell in brief about cyclic group ...i know they are abelian group or much ?

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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.