Answer to Question #203026 in Abstract Algebra for Priya

Question

Using Cayley’s theorem, find the permutation group to which a cyclic group of 

order 12 is isomorphic.

Answer 1

As given a cyclic group G = <a> of order 12 generated by the element 'a’ of G, we may take the 'permutation group’ as the cyclic subgroup H of S_12 generated by the cycle (1,2,3,4,5,6,7,8,9,10,11,12) of length 12. This (or a conjugate of it) is what we obtain as per the proof of the Cayley's Theorem.

We can also find isomorphic copies of G as subgroups of groups S_n for all n >/= 7.

E g. the product (1,2,3,4)•(5,6,7) of disjoint cycles in S_7 has order LCM(4,3)=12 and hence generates a cyclic subgroup of S_7 which is isomorphic to G.