If $G = C_{25}\times C_{45}\times C_{48}\times C_{150}$, How many subgroups of order $5$ does $G$ have?

Question

Suppose that $G = C_{25}\times C_{45}\times C_{48}\times C_{150}$, where $C_n$ denotes a cyclic group of order $n$.

How many subgroups of order $5$ does $G$ have?

I've been trying to understand this in terms of the generators of each cyclic group. We know, for example, that if $C_{45}=<c>$, that the subgroups of $C_{45}$ are $<c^3>,<c^9>,<c^5>,<c>,$ and $<1>$, so there is one subgroup of order 5.

Then we see that $C_{25},C_{45},$ and $C_{150}$ all have one subgroup of order $5$. What does that tell us about the number of subgroups of the product?

Any help appreciated!

Answer 1

You've already asked, about this same group, how many elements of order 5 it has. In particular, embedded in Mark Fischler's answer we find the answer, 124.

Now, any element of order 5 generates a subgroup of order 5. But the elements $g, g^2, g^3, g^4$ will all generate the same subgroup. So the number of subgroups is $124/4 = 31$.