# Two free basis of a free abelian group

We have a free abelian group $A(X)$, where $X$ is its free basis, and let $Y$ another free basis for $A(X)$. We know that every $g\in A(X)$ can be expressed as $g=a_1x_1+...+a_nx_n$ where the $a_i$'s are integers and $x_i$'s are pairwise distincts elements of $X$ (this is called the normal form of $g$ with respect to the basis $X$), and similarly for $Y$.

Let $x\in X$.

I want to prove that there exists $y\in Y$ such that $y=a_0x+a_1x_1+...+a_nx_n$ with $a_0\neq 0$, that is, $x$ appears in the normal form of $y$ with respect to the basis $X$.

I tried to use that $Y$ is a free basis, but I don't know how to find such a $y$.

Any hint? Thanks :)

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Take any y in Y and $a_0=1$. That will work by the definition of a basis.