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?
Best Answer
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: $$ \frac{3\cdot 8\cdot 24}{\varphi(30)}= \frac{3\cdot 8\cdot 24}{1\cdot 2\cdot 4}=72 $$