Given a free abelian group and a basis, prove there exists an automorphism between two elements

I'm studying free abelian groups and still don't have the intuition on how to approach questions.

Let $$A$$ be an abelian free group, and let {$$e_1, e_2, e_3$$} some basis. Let $$a=2e_1+7e_3$$ and $$b=3e_2+5e_3$$. Prove there exists an automorphism $$\phi$$ such that $$\phi(a)=b$$.

I do not seek for a solution, but some general intuition when facing this topic.

• abelian free groups are free $\mathbb Z$ modules. So think about $\mathbb Z$-module homomorphisms. – Don Thousand Feb 16 at 16:57

You can complete $$\{a\}$$ to a basis and $$\{b\}$$ to a basis. This is due to the fact that the coefficients in their definition are relatively prime. This is certainly a necessary condition, for if, say, $$a = d_1e_1 + d_2 e_2 + d_3 e_3$$ with $$1 < d \mid d_1, d_2, d_3$$, then any set $$S$$ containing $$a$$ of which $$a/d$$ is a linear combination, is necessarily linearly dependent: If $$a/d = \sum_{s \in S} \lambda_s s$$, then $$(d\lambda_a-1) a + \sum_{s \in S - \{a\}} d\lambda_s s = 0\,,$$ where $$d \lambda_a - 1 \neq 0$$ because $$d > 1$$. Less obvious is that it is also a sufficient condition. Constructing a basis containing $$a$$ is related to Bézout's theorem (in dimension $$2$$, it is exactly Bézout's theorem) and it is essentially the same issue as in this question: Can the determinant of an integer matrix with a given row be any multiple of the gcd of that row?.