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

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!

• Inside which group? – TokenToucan Dec 5 '15 at 16:30
• I apologize, somehow the group I was asking about wasn't in the question. – Brady Sheehan Dec 5 '15 at 16:32
• I fixed the formatting of the title, but please make sure this is really what you mean. And also, could you define $U(10)$? I don't think it's standard notation... – M Turgeon Dec 5 '15 at 16:34
• @MTurgeon thanks for correcting my formatting. I clarified my notation. – Brady Sheehan Dec 5 '15 at 16:37
• @CuddlyCuttlefish Is $\mathbb{Z}_{n}^\times$ standard notation for the group of integers less than n that are relatively prime to n under multiplication mod n? – Brady Sheehan Dec 5 '15 at 16:44

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?

• I'm looking for specifically the distinct cyclic subgroups of order 12. I understand that U(10)≅ℤ/4 but don't know how that helps me construct the subgroups. – Brady Sheehan Dec 5 '15 at 16:41
• Well, if $\langle i, j\rangle\in G\times H$ where $G$ and $H$ are finite abelian, then the order of $\langle i, j\rangle$ is the LCM of the order of $i$ and $j$. So how do you find elements of $\mathbb{Z}/6$ and $\mathbb{Z}/4$ such that the LCM of their orders is $12$? – rogerl Dec 5 '15 at 16:52
• I would consider all pairs ⟨i,j⟩ that satisfy the LCM equaling 12. I did this already for the notation of U(10) instead of $\mathbb{Z}_{4}$ and found 8 elements: |⟨i,j⟩|= |(1,3)| = |(1,7)| = |(5,3)| = |(5,7)| and |(2,3)| = |(2,7)| = |(4,3)| = |(4,7)|. – Brady Sheehan Dec 5 '15 at 17:00
• So you're done, aren't you? The first four elements are in one subgroup, and the other four are in the other. – rogerl Dec 5 '15 at 17:04
• Oh okay! I see it now. Thank you. As a way to check my work, is there a fact that tells me there will be 2 distinct cyclic subgroups of order 12? – Brady Sheehan Dec 5 '15 at 17:06

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.

• Thank you! I think I am starting to understand now. Since there are 4 generators of a subgroup of order 12 and I have 8 elements that each are of order 12, there must be 2 distinct cyclic subgroups since 8/4 = 2. – Brady Sheehan Dec 5 '15 at 17:13