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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.

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

Finally, any two bases of a free abelian group can be sent to each other by an automorphism (just as for vector spaces, the proof is the same). It corresponds to the relation between "active" and "passive" transformations in classical mechanics.

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  • $\begingroup$ Thank you very much for the quick elaborated response :) $\endgroup$ – Tamir Shalev Feb 16 at 17:22

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