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The number of generator of a finite cyclic group of order 28 is
1. 10
2. 8
3. 12
4. 14

1 Answer

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Correct Answer - Option 3 : 12

Concept:

If a cyclic group G is generated by an element 'a' of order 'n', then am 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 am € 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.

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