subgroup of $C_{16} \times C_{30}$,and C is the cyclic group (1)How many cyclic subgroups are in the group  $C_{16} \times C_{30}$,and C is the cyclic group
(2)How many subgroups are in the group $C_{16} \times C_{30}$,and C is the cyclic group
my opinion is  $C_{16} \times C_{30} \cong C_{2} \times C_{240}$.
so subgroups of $C_2$ are$C_2 $ and ${1}$,the number of subgroups of $C_{240}$ is 20 ,
so the number of subgroups is $2 \times 20 =40$
if (m,n)=1,then $C_{m} \times C_{m} \cong C_{mn}$ is cyclic group ,so ${1} \times C_{m}$ is cyclic group ,
the number of these is 20,
$C_{2}\times C_{m}$ is cyclic group only $ m=1,3,5,15$
but,  $C_{1} \times C_{2} =C_{2} \times C_{1}$
so the number of cyclic group is $3+20=23$
I don't know my answer is true or not 
 A: Hint: By the Chinese remainder theorem, $C_{16} \times C_{30} \cong C_{2} \times C_{240}$.
A: The essential fact in this case is that the group $G$ has the Klein Vierergruppe $C_2 \times C_2$ as a subgroup, resulting in the fact that $G$ has three elements of order $2$. With this in mind we will count the number of cyclic groups. First of all we have those who have odd order, there are four of them :$$C_1, C_3, C_5 \text{ and } C_{15}$$. Then there are those whose order is even with a squarefree factor $2$: there are three for each of them since we can obtain a generator by multiplying a generator of odd order with one of the tree elements of order $2$, this gives us twelve groups: $$C_2, C_6, C_{10}\text{ and } C_{30} $$ Let $a,b$  be the respective generators of $G = C_2 \times C_{240}$ then the remaining cyclic subgroups are the groups which order have order with non squarefree factor  $2$, there are two of each of them : those who are a subgroup of $C_{240}$ and those obtained by the previous ones by multiplying their generator with $a$ this gives $24$ cyclic subgroups : $$ C_4, C_8, C_{12},C_{16},C_{20},C_{24}, C_{40},
C_{48}, C_{60}, C_{80},C_{120}\text{ and }  C_{240}$$ As a conclusion : we have $4 + 12 + 24 = 40$ cyclic subgroups. The non cyclic subroups are easier to count since the are the direct products $C_2 \times H$ where $H$ is a subgroup of $C_{240}$ with odd order so there are $$16 of them.
Conclusion : $G$ has $40+16 = 56$ subgroups
