It's a well known fact that the number of index $n$ sublattices of a rank two lattice $\Lambda$ is given by $\sigma_1(n) = \sum_{d\mid n} d$.

Here is a proof of this fact:

Proof: **choosing a basis** of $\Lambda' \subset \Lambda$ we are really counting the number of $2 \times 2$ matrices up to right-multiplication by $SL_2^\pm\mathbb{Z}$ (we have to allow determinant $\pm1$). So let $\begin{pmatrix}a & b \\ c & d \end{pmatrix}$ be the given matrix. Furthermore, let $\gamma = \gcd(c,d)$ and let $r, s$ be such that $rc + sd = \gamma$. Then one can easily verify that
$$
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
\underbrace{\begin{pmatrix}
d/\gamma & r \\ -c/\gamma & s
\end{pmatrix}}_{\in SL_2\mathbb{Z}}
=
\begin{pmatrix}
n/\gamma & ra + sb \\ 0 & \gamma
\end{pmatrix}
$$
Furthermore, repeated post-multiplication with the matrix $\big(\begin{smallmatrix}1 & \pm1 \\ 0 & 1\end{smallmatrix}\big)$ will yield a matrix of the form
$$
\begin{pmatrix}
n/\gamma & t \\ 0 & \gamma
\end{pmatrix}
$$
with $0 \leq t < n/\gamma$. It follows that we can write our basis for $\Lambda'$ uniquely in this form; so the number of such lattices is the number of such matrices, which is clearly $\sigma_1(d)$.

One can also start by **choosing** a splitting of the lattice $0 \to \Lambda_1 \to \Lambda \to \Lambda_2 \to 0$ and looking at how $\Lambda'$ intersects with this splitting. However, morally this seems to be pretty much the exact same proof, and it fundamentally still involves a choice of the splitting.

Is there a nice proof of this fact that doesn't involve some non-canonical choices?

choice. You are counting the number of subgroups of $\mathbb{Z}^2$ of index $n$, and you observe that this corresponds to the number of $\mathrm{SL}_2^\pm\mathbb{Z}$-equivalence classes of matrices in $\mathrm{GL}_2\mathbb{Z}$ of determinant $\pm n$. (I don't see the choice yet.) Then you compute that this number is $\sigma_1(n)$, by coming up with a "normal form" for each of these equivalence classes. Is it this last step that you interpret as a "choice"? $\endgroup$ – Tom De Medts Feb 1 '16 at 15:29