A generalization: Let $G$ be a free abelian group of rank $n$, and let $H$ be a subgroup of $G$ such that $G/H$ is a cyclic group of finite order $m$. Then there is a basis $x_{1},\ldots,x_{n}$ of $G$ such that $x_{1},\ldots,x_{n-1},mx_{n}$ is a basis for $H$. Since the index of $H$ in $G$ is $m,\ H$ is also free of rank $n$, and a basis $f_{1},\ldots,f_{n}$ of $H$ is obtained from some basis $e_{1},\ldots,e_{n}$ of $G$ through a linear transformation $A$ with integral coefficients and determinant $1$. By the Invariant Factor Theorem, there are invertible integral matrices $P,Q$ such that $PAQ$ is diagonal, and its diagonal entries are $d_{1}|d_{2}|\ldots|d_{n}$ with $\prod d_{i}=m$ (we can adjust $P,Q$ such that $d_{i}\ge 1$, of course). The exponent of $G/H$ is thus $d_{n}$, and since we know that it is $m$, we must have $d_{1}=\ldots=d_{n-1}=1$, and $d_{n}=m$. This finishes the proof. P.S. I hope I understood the problem correctly. I don't know what $A(K), A(L)$ are, but I interpreted $\frac{A(K)}{A(L)}=m$ as $[G: H]=m$, i.e. the same thing as saying that $H$ has index $m$ in $G$, and the next statement about $m$ as saying that $G/H$ has exponent $m$. This information is enough to conclude that $G/H$ is cyclic of order $m$.