How to prove that $\frac{1}{n}L/L\simeq (\mathbb{Z}/n\mathbb{Z})^2$? Let $L$ be any lattice in $\mathbb{C},$ and $L'$ a lattice containing $L$ with index $n$ (i.e $n=\sharp L'/L$)  I found this statement 
"The lattice $L'$ must  be contained in  $\frac{1}{n}L = \{\frac{1}{n}L\; ; \; l\in L\},$ because each element of $L'/L$ has order dividing $n=\sharp L'/L.$ Thus, $L'$  corresponds to a subgroup of order $n$ in $\frac{1}{n}L/L\simeq (\mathbb{Z}/n\mathbb{Z})^2$
"
in  Koblitz Introduction to Elliptic Curves and modular forms  p 152
I don't understand why


*

*$L'\subset \frac{1}{n}L$

*$L'$ corresponds to a subgroup of order $n$ in
$\frac{1}{n}L$

*$\frac{1}{n}L/L\simeq (\mathbb{Z}/n\mathbb{Z})^2$


Could someone please explain to me this, Thanks in advance.
 A: When I was first looking at questions like this I recall needing to take my time to fully absorb the fact that $\frac1n L$ is BIGGER than $L$. Why would anyone think that multiplication by $1/n$ makes something smaller? Beats me:-) Anyway, with that off my chest, let's roll.
1. Let $x\in L'$. Consider the coset $x+L$ in the quotient group $L'/L$. Because $[L':L]=n$ we have, by Lagrange's theorem familiar from elementary group theory, that $$nx+L=n(x+L)=0_{L'/L}=L.$$ So $nx\in L$ and therefore $x\in\frac1n L$.
2. By part 1 we have the inclusions
$$
L\subseteq L'\subseteq \frac1nL.
$$
Moding out the smallest subgroup $L$ tells us that
$$
L'/L\le \frac1nL/L.
$$
We were given the fact that $\#L'/L=n$ so the subgroup $L'/L$ has, indeed, order $n$. In other words, we are using the fact that if $K$ is a normal subgroup of $G$, then there is a bijective correspondence between the subgroups of $G/K$ and the intermediate groups $H$ such that $K\le H\le G$. The correspondence relates $H$ and $H/K\le G/K$.
3. $L$ is a free abelian group of rank two generated by, say, $\alpha$ and $\beta$. So $\frac1n L$ is the free abelian group generated by $u=\alpha/n$ and $v=\beta/n$. We then switch points of view and think of $L$ as the subgroup of $\frac1nL$ generated by $nu$ and $nv$. The isomorphism is given by
$$
f:\frac1nL/L\to(\Bbb{Z}/n\Bbb{Z})^2, (au+bv)+L\mapsto (\overline{a},\overline{b}).
$$
A: Let $L'=1/nL$, $(1/nL)/L=L'/nL'= Z^2/nZ^2=(Z/nZ)^2$
