How many number of cyclic subgroup of order $12$ of $\mathbb Z/2\mathbb Z×\mathbb Z/3\mathbb Z×\mathbb Z/3\mathbb Z ×\mathbb Z/4\mathbb Z$?

Question

How many number of cyclic subgroup of order $12$ of $\mathbb Z/2\mathbb Z×\mathbb Z/3\mathbb Z×\mathbb Z/3\mathbb Z ×\mathbb Z/4\mathbb Z$ ?

I think first try is to count the number of element of order $12$, and that is $2\times 3\times 3\times 2=36$ (element of order of $\mathbb Z/4\mathbb Z$ need to be 4).

The number of element of order $12$ and the number of subgroup of order $36$ may be related, I think I need to divide $36$ by some number .

Is my attempt right? And how can I correctly deal with it ?

Thank you in advance.

Answer 1

Let $G=C_2\times C_3\times C_3\times C_4$ (here, $C_n$ is the cyclic group of order $n$.) By the Chinese Remainder Theorem, the cyclic subgroups $A\subseteq G$ of order $12$ correspond to pairs $(B,C)$ of a cyclic subgroup $B\subseteq G$ of order $3$ and a cyclic subgroup $C\subseteq G$ of order $4$. (given by $A\mapsto (3A,4A)$ and $(B,C)\mapsto B\times C$, the inner direct product).

Now, the order $3$ cyclic subgroups $B\subseteq G$ are just order $3$ subgroups of $C_3\times C_3$, which is simple to count. Similarly, the order $4$ cyclic subgroups $B\subseteq G$ are just order $4$ subgroups of $C_2\times C_4$.