Existence of finitely presented group with exactly $k$ subgroups?

Question

Given any natural number $k$, does there necessarily exist a finitely presented group $G$ with $k$ subgroups? I'm trying to show that having $k$ subgroups is a markov property, but google doesn't seem to yield much on this matter.

Answer 1

As you didn't say non-Abelian, I snipe the cheepo: $\mathbb{Z}_{p^{k-1}}$ for any prime $p$ has exactly $k$ subgroups, as $p^{k-1}$ has exactly $k$ divisors.

Answer 2

We can even take a finite abelian group, namely the cyclic group $\Bbb{Z}/p^{k-1}$. This site also has the question to find a non-abelian group with exactly $k$ normal subgroups.

Reference: Non-abelian finite groups with exactly $n$ normal subgroups.