I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.39.
Give an example of a group that has exactly $6$ subgroups (including the trivial one and the group itself). Generalise to exactly $n$ subgroups for any positive integer $n$.
My Attempt:
I suspect that $\Bbb Z_{p^3q}$ has six subgroups for distinct primes $p$ and $q$. Why? Because I can sort of visualise its subgroup Hasse diagram as two parallelograms of the same size, sharing an edge of the same size, with exactly six vertices/nodes on the diagram corresponding to the subgroups.
The above is vague. I'm not interested in computing the subgroups (yet) as I doubt doing so would be edifying.
So I can't say for sure whether my candidate group works.
The solutions in the book are as follows (and I quote):
"For $6$, use $\Bbb Z_{2^5}$. For $n$, use $\Bbb Z_{2^{n-1}}$."
My main problem:
How does one go about answering Exercise 4.39 using the tools available in the book so far, without doing too many calculations?
Please help.
Hint: If $G $ is a cyclic group of order $n$, then for each positive integer $m$ dividing $n$ there is a unique subgroup of $G$ of order $m $. How many divisors are there for $2^{n-1}$?