Elements and cyclic subgroups of order $15$ in $\Bbb Z_{30}\times \Bbb Z_{20}.$ This is Exercise 8.22 of Gallian's, "Contemporary Abstract Algebra". Please use only methods from this book prior to the exercise.
This is an alternative-proof question.

Find the number of elements of order $15$ and the number of cyclic subgroups of order $15$ in $\Bbb Z_{30}\times \Bbb Z_{20}.$

Thoughts:
This exercise gives two distinct tasks, yes, but I think it's to highlight the discrepancy between the two answers.
I've done some work below in GAP for this but I don't consider it a solution within the spirit of the text.
gap>  F:=FreeGroup(2);
<free group on the generators [ f1, f2 ]>
gap> 
gap> rels:=[(F.1)^(30), (F.2)^(20), (F.1)*(F.2)*(F.1)^-1*(F.2)^-1];
[ f1^30, f2^20, f1*f2*f1^-1*f2^-1 ]
gap> G:=F/rels;
<fp group on the generators [ f1, f2 ]>

This sets the group up, calling it G.
gap> S:=[];
[  ]
gap> 
gap> for g in G do
> if Order(g)=15 then
> AddSet(S, g);
> fi;
> od;
gap> 
gap> 
gap> Size(S);
48

So I think there are $48$ elements of order $15$ in G.
gap> C:=[];
[  ]
gap> 
gap> for g in G do
> H:=Subgroup(G, [g]);
> if Size(H)=15 then
> AddSet(C, H);
> fi;
> od;
gap> 
gap> 
gap> Size(C);
6

So I think there are $6$ cyclic subgroups of G of order $15$.
That's all I have.

I think I should be able to do this exercise myself but I lack the time to take it any further without help.
Here is a similar question I asked: Is $|\{z\in\Bbb Z_3\times\Bbb Z_9: |z|=9\}|=18?$

Please help :)
 A: I don't have this text, so I don't know which material is available to you, but the following only uses basic facts about cyclic groups and coprimality.
Hint Since $\Bbb Z_{20} \cong \Bbb Z_4 \times \Bbb Z_5$ and $\Bbb Z_{30} \cong \Bbb Z_2 \times \Bbb Z_3 \times \Bbb Z_5$, we have $$\Bbb Z_{20} \times \Bbb Z_{30} \cong \Bbb Z_2 \times \Bbb Z_3 \times \Bbb Z_4 \times \Bbb Z_5^2 .$$
Since the group is abelian, any element of order $15$ is a product of an element of order $3$ and an element of order $5$; conversely, any such product has order $15$ and any two such products are distinct. How many elements of order $3$ are there in the above product? Of order $5$?
With that number in hand, each element of order $15$ is contained inside exactly one subgroup of order $15$ (namely, the one it generates). How many elements of order $15$ are in each such subgroup?

From the above decomposition we see that there are two elements of order $3$ and $24$ elements of order $5$, so there are $2 \cdot 24 = 48$ elements of order $15$. Now, any subgroup of order $15$ contains $\phi(15) = 8$ elements of order $15$, and any element of order $15$ is in precisely one subgroup of that order (namely the one it generates), so the $48$ elements of order $15$ are evenly divided across $\frac{48}{8} = 6$ subgroups of that order.

