Correct Answer - Option 3 : 12

**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, a^{2}, a^{3}, ........ , a^{28} = 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, a^{3}, a^{5}, a^{9}, a^{11}, a^{13}, a^{15}, a^{17}, a^{19}, a^{23}, a^{25}, a^{27 }are the generators of G.

Therefore, there are **12 **generators of G.