Number of subgroup of a group $G$ with order 2021 OK so here's the question I stumbled upon

Let $G$ is a group with order 2021, how many subgroup does G have?

the thing is, I personally think that the question itself lacks additional info, so I'm not sure on how should I approach it.
So far I've been thinking that, we know $2021$ is not a prime number and that the factors of $2021$ are
$1, 43, 47, 2021$
by assuming it's cyclic, the number of subgroup is $2^k$ with $k$ is the number of distinct factors of $2021$, then we have
$2^k=2^2=4$
is this correct? any insight would really helps, thank you.
 A: Note that $2021 = 43 \cdot 47$ and $43 \not\mid (47-1)$ hence classification theorems give $G \simeq \mathbb{Z}_{2021}$. (If you like by chinese remainder theorem you also have $\mathbb{Z}_{2021} \simeq \mathbb{Z}_{43} \times \mathbb{Z}_{47}$).
You reduced the problem to the studying of the number of subgroups of a cyclic group, you can find references in the following : Determine the number of subgroups of a cyclic group, Finding all the subgroups of a cyclic group, Prove that the total number of subgroups of a finite cyclic group $G=\langle a \rangle$ such that $o(G)=n$ is the number of divisors of $n$.
A: $|G|=43*47$
By Sylow Theorem $\exists S$ $47-Sylow$ and $\exists T$ $43-Sylow$. Lets call $n_{47}$,$n_{43}$ the number of $47-Sylow$ and $43-Sylow$.
$n_{47}= 1 mod (47)$ so $n_{47}=47\lambda+1$ and $n_{47}|(G:S)$ so appliying Langrange theorem you have that $n_{47}|43$ but $n_{47}=47\lambda+1$ so $\lambda=0$ and $n_{47}=1$
In an analogous way we have that $n_{43}=1$.
There are two unique sylows $S$ and $T$, $|G|=|S||T|$ and $S\cap T=\{1\}$ so $G\cong S\times T$ and for this abelian( Because $S$ and $T$ are) and cyclic ($43$ and $47$ coprimes), this is little redundant because clyclic implies abelian but anyway. So $G\cong \mathbb{Z}_{2021}$
and it has as subgoups ( whithout counting isomorphims) as divisor of 2021.
Just remember that by Lagrange theorem consecuences if $H<G$ then $|H|||G|$.
Controversly when $G$ cyclic and $|G|=n$ if $d|n$ then $\exists H<G$ such that $ |H|=d$. Don´t forget that every cyclic group $|H|=d$ is isomorphic.
So your problem is now reduced to calculate 2021 divisors number.
