What happen when a lattice is contained in its dual? Im reading this paper by J.H Conway about coding theory which involves basic knowledge on lattices. 
In the document they say that a lattice $\Lambda :=\{ u_1\vec{a}_1+\cdots+u_m\vec{a}_m| u_i \in \mathbb{Z} \} $ is not always contained in its dual $ \Lambda^*:=\{ y \in span(\vec{a}_1,\cdots,\vec{a}_m)| \langle x,y\rangle \in \mathbb{Z}\; \forall x \in \Lambda  \}$ but when it is the case that $ \Lambda \subseteq \Lambda^*$ there exists $r_0, \cdots, r_{d-1} $ vectors such that $$  \Lambda^*=\underset{i=0}\bigcup^{d-1} \left( r_i+\Lambda \right) $$
Since there is not an explanation of this fact in the paper I think that it would be easy to figure out why it is true but I'm still stuck on it. 
Can someone give me a clue or at least more documentation on this topic in which this could have a more detailed explanation? 
Thank you all
 A: What's going on is the following:


*

*The lattice $\Lambda$ is a free abelian group of rank $m$.

*The lattice $\Lambda^*$ is another free abelian group of rank $m$ (to get that we need the restriction $y\in\operatorname{span}(\vec{a}_1,\vec{a}_2,\ldots,\vec{a}_m)$).

*$\Lambda\subseteq \Lambda^*$.


These imply (for example by the structure theory of finitely generated abelian groups) that the quotient group $\Lambda^*/\Lambda$ is finite. So $d$
is the order of this group, and $r_0,r_1,\ldots,r_{d-1}$ are the coset representatives. It is, indeed, the case that $d$ can be calculated as that determinant.

All these facts are relatively basic facts about the algebra/geometry of lattices in a Euclidian space. Not sure where to refer you to. The Wikipedia article on lattices seems singularly useless. It jumps from 2D/3D-lattices to Haar measures fer crying out loud. Nothing about the Gram matrix. The WP article on Gram matrices may be more informative, but doesn't really explain the connection with the order of the quotient group. It is "sort of" there if you use Haar measures to calculate the index :-(
