Cyclic subgroups of maximum possible order of $\Bbb Z_6\times\Bbb Z_{10}\times\Bbb Z_{15}$ of the form $⟨a⟩\times⟨b⟩\times⟨c⟩.$ I was doing problems from Gallian and I found the following one:
Find three cyclic subgroups of maximum possible order of $\mathbb Z_6\times \mathbb Z_{10}\times \mathbb Z_{15}$ of the form $\langle a \rangle \times\langle b \rangle \times \langle c \rangle$ where $a,b,c$ are members of the $3$ component groups respectively.
Soln: The maximum possible order of a cyclic subgroup is $\mathbb{lcm}(6,10,15)=30$.
Now, we can have the cyclic subgroups of  $C_2\times C_5\times C_3$  and $C_3\times C_2\times C_5$ and $C_6\times \{e\}\times C_5$ .
which are $\langle 3 \rangle\times \langle 2 \rangle\times \langle 5 \rangle$ and $\langle 2 \rangle\times \langle 5 \rangle\times \langle 3 \rangle$ and $\langle 1 \rangle\times\langle 0 \rangle\times \langle 3 \rangle$.
There are other cyclic subgroups too,for example $C_2\times \{e\}\times C_{15}$ is obtained by,$\langle 3 \rangle\times \langle 0 \rangle\times\langle 1 \rangle$.
Is my solution correct?What is the complete collection of such cyclic subgroups and how can I determine how many are there?
 A: Yes, what you did is correct, but you missed some subgroups.
The idea is to note that the subgroups you are looking for are of the form:
\begin{equation}
C_x\times C_y\times C_z
\end{equation}
and you want
\begin{equation}
C_x\times C_y\times C_z \cong C_{30}
\end{equation}
with $x\in\{1,2,3,6\}$, $y\in\{1,2,5,10\}$ and $z\in\{1,3,5,15\}$. You can take $x,y,z$ in the set of all respective divisors because the groups $\mathbb Z_6$, $\mathbb Z_{10}$, $\mathbb Z_{15}$ are cyclic.
Thanks to the Chinese remainder theorem the problem is equivalent to find triple $(x,y,z)$ (taken in the sets above) such that $xyz=30$. By a direct computation we get $8$ triples:
\begin{gather}
(1,2,15)\\
(1,10,3)\\
(2,5,3)\\
(2,1,15)\\
(3,10,1)\\
(3,2,5)\\
(6,5,1)\\
(6,1,5)
\end{gather}
That correspond to all the subgroups you are looking for.

The number of cyclic subgroups of order $30$ in $G$ is bigger. To calculate this number is sufficient to count all the element of order $30$ in $G$ and divide this number by $\varphi(30)$ because every cyclic subgroup of order $30$ has exactly $\varphi(30)$ generators.
Since $G\cong \mathbb Z_2\times\mathbb Z_2\times\mathbb Z_3\times\mathbb Z_3\times\mathbb Z_5\times\mathbb Z_5$ the number of element of order $30$ is
$$
(2^2-1)(3^2-1)(5^2-1) = 3\cdot 8\cdot 24
$$
So the number of cyclic subgroups is:
$$
\frac{3\cdot 8\cdot 24}{\varphi(30)}= \frac{3\cdot 8\cdot 24}{1\cdot 2\cdot 4}=72
$$
A: You did it correctly.  To get maximal cyclic subgroups: $\Bbb Z_2×\Bbb Z_5×\Bbb Z_3,\Bbb Z_3×\Bbb Z_{10}×e,\Bbb Z_2×e×\Bbb Z_{15}$ will work.  All three are isomorphic to $\Bbb Z_{30}$.
