# 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?

• Use $\langle x\rangle$ for $\langle x\rangle$. – Shaun Apr 29 '20 at 15:33
• @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. – Kishalay Sarkar Apr 29 '20 at 15:37
• Fair enough. Sorry. – 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: $$$$C_x\times C_y\times C_z$$$$ and you want $$$$C_x\times C_y\times C_z \cong C_{30}$$$$ with $$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 $$\Bbb Z_{30}$$.

• thanks..................................... – Kishalay Sarkar Apr 29 '20 at 15:17