Dimension of span of a finitely generated subgroup

Question

If $QA\subset V$ is the span of $A\subset V$, V- a vector space over the field $Q$, $A$ - a finitely generated abelian subgroup (and hence has a finite rank), then is $dim(QA) = rank(A)?$ I think since $A$ spans $QA$, any basis of $A$ has a corresponding basis in $QA$ and hence the statement is true? Intuitively, it feels right but not sure about concrete reason. A hint is appreciated.

Answer 1

Let $\{e_1,\dots,e_k\}$ be a basis in $A$. Then it is also a basis in $QA$. Indeed, every element of $QA$ is a rational linear combination of elements of $A$ and hence a rational linear combination of $e_1,\dots,e_k$. Let us prove that $e_1,\dots, e_k$ are linearly independent in $QA$. Assume that they are, $\lambda_1 e_1+\dots+\lambda_ke_k=0$. Let $q$ be a common denominator of all nonzero $\lambda_j$, and let $m_j=q\lambda_j$. Then all $m_j$ are integers and $m_1e_1+\dots+m_ke_k=0$, and hence all $m_j=0$ by the linear independence of $e_1,\dots,e_k$ in $A$.