What are the distinct cyclic subgroups of order 12 in $\mathbb{Z}_6 \times \mathbb{Z}_{10}^\times$?

Question

I am trying to find the cyclic subgroups of order 12 in the group $\mathbb{Z}_6 \times\mathbb{Z}_{10}^\times$.

I know that there will be 4 cyclic subgroups of order 12 by the euler phi function and some of them are (1,3), (1,7), (5,3), and (5,7). But the groups generated by these elements are identical. What are the other cyclic subgroups?

In addition, I have in my notes that there are two distinct cyclic subgroups of order 12. How do I know that there are two distinct cyclic subgroups?

Thanks!

Answer 1

This problem might be easier to think about if you recognize that $U(10) \cong \mathbb{Z}/4$ (via, for example, the isomorphism given by extending $3\mapsto 1$). Then the group you give is isomorphic to $\mathbb{Z}/6\times\mathbb{Z}/4$. Now can you figure out how to construct order $12$ subgroups?

Answer 2

1,3,7,11 are co-prime to 10 and 1,5 are co prime to 6.So, there are 4*2=8 elements of order 12 in Z6×Z×10.Now,each cyclic subgroup of order 12 contains 4 elements of order 12 and there are only 8 elements of order 12 in the given group, so there must be 8/4=2 cyclic-subgroups of order 12.