Let $\Lambda$,$\Lambda'$ be two lattice in $\mathbb{C}$ and $m\neq 0\in\mathbb{C}$ satisfying $$ m\Lambda\subset\Lambda' $$
The, the book I'm reading says that by the theory of finite Abelian groups there exists a basis $\{\omega_1,\omega_2\}$ of $\Lambda$' and positive integers $n_1,n_2$ such that $\{n_1\omega_1,n_2\omega_2\}$ is a basis of $m\Lambda$.
I wonder which theorem is it using to deduce such conclusion? There is even no finite Abelian groups there.
update:
I think the author means theory of finitely-generated Abelian groups, doesn't him?
Pick a basis $\lambda_1, \lambda_2$ for $\Lambda$ and a basis $\mu_1, \mu_2$ for $\Lambda'$. By hypothesis we may write $m\lambda_i$ as a linear combination of $\mu_1$ and $\mu_2$. Write the matrix $$ \pmatrix{a & b \\ c & d } $$ where the $i$th column is $m\lambda_i$ expressed in terms of the basis $\mu_i$. You are allowed to do $\mathbb{Z}$-linear row and column operations to this matrix. Row operations correspond to changing the basis for $\Lambda'$, and column operations correspond to changing the basis for $\Lambda$. Because $\mathbb{Z}$ is a PID, you can make this matrix diagonal with those operations (same argument as in proof of fundamental theorem for finitely generated modules over a PID). That exactly says there that the bases you seek for $\Lambda$ and $\Lambda'$ exist (and gives you an algorithm to find them, if you keep track of the row and column operations).