The number of generator of a finite cyclic group of order 28 is

Concept:

If a cyclic group G is generated by an element 'a' of order 'n', then a^m is a generator of G if m and n are relatively prime.

Calculation:

Let a cyclic group G of order 28 generated by an element a, then

o(a) = o(G) = 28;

to determine the number of generator of G evidently,

G = {a, a², a³, ........ , a²⁸ = e}

An element a^m € G is also a generator of G is H.C.F of m and 28 is 1.

H.C.F of (1, 28) is 1 silimarly, (3, 28) (5, 28), (9,28) (11, 28), (13, 28), (15, 28), (17, 28), (19, 28), (23, 28), (25, 28), (27, 28)

Hence a, a³, a⁵, a⁹, a¹¹, a¹³, a¹⁵, a¹⁷, a¹⁹, a²³, a²⁵, a²⁷ are the generators of G.

Therefore, there are 12 generators of G.