# The number of cyclic subgroups of order 15 in $\mathbb{Z}_{30} \oplus \mathbb{Z}_{20}$

What is the number of cyclic subgroup of order 15 in $\mathbb{Z}_{30} \oplus \mathbb{Z}_{20}$?

I have counted the number of elements with order 15 in $\mathbb{Z}_{30} \oplus \mathbb{Z}_{20}$ that is $48$, by counting all possibilities of $(a,b)$ where $a\in\mathbb{Z}_{30}$ , $b \in \mathbb{Z}_{20}$ and $\def\lcm{\operatorname{lcm}}\lcm(|a|,|b|)=15$.

My argument is that for an element that has order $15$, say $(x,y)$, then

$\langle(x,y)\rangle = \langle(x,y)^2\rangle = \langle (x,y)^4\rangle = \langle (x,y)^7\rangle = ...=\langle (x,y)^{14}\rangle$

So there are $8$ different elements with order $15$ such that the generate the same subrgroup. So my answer is that there are $\frac{48}{8}=6$ different cyclic subgroups of order 15, am i correct? or maybe there are more simpler ways to solve this problem?

• Do you know about the structure theorem for finitely generated Abelian groups? – Marc van Leeuwen Nov 11 '13 at 16:35
• @MarcvanLeeuwen no i have'nt study that yet. My answer seems correct right? since every $8$ elements among the $48$ elements of order $15$ generate the same subgroup – Arief Anbiya Nov 11 '13 at 16:38

## 1 Answer

Provided you correctly counted the elements of order$~15$, your answer is correct. You can indeed count cyclic subgroups by counting their generators (elements or order$~n$) and dividing by the number $\phi(n)$ of generators per cyclic subgroup, since every element of order$~n$ lies in exactly one cyclic subgroup of order$~n$ (the one that it generates).

Here is how I would count the elements of order$~15$. By the Chinese remainder theorem one has $\newcommand\Z{\Bbb Z_{#1}}\Z{30}\cong\Z2\oplus\Z3\oplus\Z5$ and $\Z{20}\cong\Z4\oplus\Z5$, so all in all we are dealing with the group $\Z2\oplus\Z4\oplus\Z3\oplus\Z5^2$. To have order $15$, an element must have a trivial (zero) component in $\Z2$ and $\Z4$, in $\Z3$ it must have as component one of the $2$ generators, and it's component in $\Z5^2$ must be any one of the $24$ nonzero elements. Indeed you get $2\times24=48$ elements of order$~15$.