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?

  • $\begingroup$ Use $\langle x\rangle$ for $\langle x\rangle$. $\endgroup$ – Shaun Apr 29 '20 at 15:33
  • $\begingroup$ @Shaun see that I have used it in the body of question,but I could not do it in title because it would exceed the word limit. $\endgroup$ – Kishalay Sarkar Apr 29 '20 at 15:37
  • $\begingroup$ Fair enough. Sorry. $\endgroup$ – Shaun Apr 29 '20 at 15:39

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


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 $\Bbb Z_{30}$.

  • 1
    $\begingroup$ thanks..................................... $\endgroup$ – Kishalay Sarkar Apr 29 '20 at 15:17

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