Let $R$ be a $M\times N$ matrix with rational entries. Is $|(R\mathbb{Z}^N)/\mathbb{Z}^M|=|(R^T\mathbb{Z}^M)/\mathbb{Z}^N|$? Let $R$ be a $M\times N$ matrix with rational entries, $R\mathbb{Z}^N$ be the image of $\mathbb{Z}^N$ under $R$.
Consider an equivalence relation on $R\mathbb{Z}^N$ defined by
$a\sim b$ if $a-b\in \mathbb{Z}^M$ for any $a,~b\in R\mathbb{Z}^N$.
Denote the set of equivalent classes as $(R\mathbb{Z}^N)/\mathbb{Z}^M$.
Similarly, we have the notion of $(R^T\mathbb{Z}^M)/\mathbb{Z}^N$.
Both $(R\mathbb{Z}^N)/\mathbb{Z}^M$ and $(R^T\mathbb{Z}^M)/\mathbb{Z}^N$ form groups under addition.
Questions:
(1) Is $(R\mathbb{Z}^N)/\mathbb{Z}^M$ isomorphic to $(R^T\mathbb{Z}^M)/\mathbb{Z}^N$?
(2) If (1) is not true,  is the cardinality $|(R\mathbb{Z}^N)/\mathbb{Z}^M|=|(R^T\mathbb{Z}^M)/\mathbb{Z}^N|$?
(This is posted on both Math Overflow and Math Stack Exchange.)
 A: First of all we can multiply all entries of $R$ by a positive integer $k$ such that $kR\in M_{m\times n}(\mathbb Z)$. (In the following I'll change your notation from $M$ to $m$, and $N$ to $n$.) Since $R\mathbb Z^n/\mathbb Z^m\simeq kR\mathbb Z^n/k\mathbb Z^m$ we can reduce the problem to the following:

Let $R\in M_{m\times n}(\mathbb Z)$ and $k$ a positive integer. Are $R\mathbb Z^n/k\mathbb Z^m$ and $R^T\mathbb Z^m/k\mathbb Z^n$ isomorphic?

The Smith normal form of $R$ has the invariant factors $d_1,\dots,d_r$. Thus we have $R\mathbb Z^n=\mathbb Zd_1x_1\oplus\cdots\oplus\mathbb Zd_rx_r$ for some basis $\{x_1,\dots,x_m\}$ of $\mathbb Z^m$. Then $R\mathbb Z^n\cap k\mathbb Z^m=(\mathbb Zd_1\cap\mathbb Zk)x_1\oplus\cdots\oplus(\mathbb Zd_r\cap\mathbb Zk)x_r$ and therefore $$R\mathbb Z^n/R\mathbb Z^n\cap k\mathbb Z^m\simeq \mathbb Zd_1/(\mathbb Zd_1\cap\mathbb Zk)\oplus\cdots\oplus\mathbb Zd_r/(\mathbb Zd_r\cap\mathbb Zk).$$
Since the invariant factors of $R^T$ are also $d_1,\dots,d_r$ (see here) we get $$R\mathbb Z^n/R\mathbb Z^n\cap k\mathbb Z^m\simeq R^T\mathbb Z^m/R^T\mathbb Z^m\cap k\mathbb Z^n.$$
But from an isomorphism theorem we know that $R\mathbb Z^n/k\mathbb Z^m\simeq R\mathbb Z^n/R\mathbb Z^n\cap k\mathbb Z^m$, so we have proved that the two $\mathbb Z$-modules are isomorphic. 
