There are only finitely many integer lattices with bounded covolume I have an uninsightful proof for the following lemma (I discuss motivation below). Let $C>0$ and $n\geq 1$ be fixed, set $\def\Z{\mathbb{Z}}M=\Z^n$.

Lemma. There are only finitely many subgroups $H\subset M$ with $\#(M/H) \leq C$.

Below is a slightly sloppy write-up of it.
Proof. There are finitely many isomorphism classes of abelian groups of order $\leq C$, say there are $N$ of them. For every isomorphism class pick out one representative, e.g. $R_i$ for $i=1,\dots,N$. Let $H\subset M$ be a lattice with $\#(M/H)\leq C$ and pick an isomorphism $\varphi_H:M/H\xrightarrow{\,\sim\,} R_i$ for the appropriate $i$. Choose a fixed basis $(e_1,\dots,e_n)$ of $M$ and consider the $n$-tuple $\big(\varphi_H(\pi_H(e_1)),\dots,\varphi_H(\pi_H(e_n))\big)$ where $\pi_H:M\to M/H$ is the canonical projection. We have thus associated to $H$ a tuple
$$
\Big(\varphi_H(\pi_H(e_1)),\dots,\varphi_H(\pi_H(e_n)\Big)
\in
\coprod_{i=1}^NR_i^n
$$
Clearly if $H$ and $H'$ are mapped to the same tuple the kernels of $\varphi_{H}\circ\pi_H$ and $\varphi_{H'}\circ\pi_{H'}$ are equal i.e. $H=H'$. There are therefore at most
$$
N C^n
$$
sublattices of $M$ with covolume $\leq C$.
Question

Is there a better, more insightful, proof of this statement? Is there a good reference for this?

My initial thought was to use short bases: there are only finitely many short bases (as there are only finitely many short vectors) so proving that any lattice with bounded covolume has a short basis (all vectors of length $\leq$ than some constant depending only on $n$ and $C$) would do the trick. I have a passing familiarity with some of the result in this field (Minkovski's inequality to produce short vectors, LLL's algorithm to produce short bases) and I expect they provide a better bound on the number of lattices with bounded covolume.
Motivation
The reason I'm asking this question is that this seemed like a natural way to prove finiteness of class groups of number fields $K/\Bbb{Q}$ given the result that any ideal class contains an integral ideal $\mathfrak{b}$ with
$$
N(\mathfrak{b})\leq\left(\frac4\pi\right)^{r_2}
\frac{n!}{n^n}\sqrt{d}
$$
where $d$ is the absolute discriminant of $K$. The proof I know uses other facts about ideal containment to deduce that there are only finitely many ideals with given norm. The result above seemed like it should be true and an alternative way to finish the proof of finiteness of class groups.
 A: I think your proof is fine. FWIW it's possible to count exactly how many sublattices there are of a fixed index. I think the proof can be done using Smith normal form (edit: actually I guess Hermite normal form is what we really want), although you can see this blog post for a funny alternative proof using generating functions. The answer is that the number of sublattices of $\mathbb{Z}^n$ of index $D$ is given by
$$\sum_{d_1 d_2 \dots d_n = D} d_1^{n-1} d_2^{n-2} \dots d_{n-1}.$$
When $n = 2$ this is the divisor function $\sigma_1(D)$. So the number of sublattices of index $\le D$ in this case is
$$\sum_{i=1}^D \sigma_1(i) \sim D \log D + O(D)$$
whereas your bound grows at least quadratically in $D$ but is complicated by needing to count isomorphism classes of abelian groups.
Generally, as a function of $D$ this is a multiplicative arithmetic function with Dirichlet series $\zeta(s) \zeta(s - 1) \dots \zeta(s - (n-1))$. Presumably a generalization of the Dirichlet hyperbola method can be used to find asymptotics. The largest term occurs when $d_1 = D$ and gives $D^{n-1}$ so maybe the asymptotics look like $D^{n-1} (\log D)^r$ for some $r$ which may be a function of $n$?
