Using Cayley’s theorem, what is the permutation group to which a cyclic group of order 12 is isomorphic to?
Answer:-
We have given cyclic group G = <a> of the order 12 generated by the element 'a’ of G,
we will take the' permutation group as we we have the cyclic subgroup H of S_12 generated by the cycle , that is (1,2,3,4,5,6,7,8,9,10,11,12) of the length 12 and
This (or a conjugate of it) is what we obtain as by the proof of the Cayley's Theorem.
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