Let $A,B,C$ be finitely generated abelian groups. Assume that there is an exact sequence $$0 \to C \to A^n \to B^n \to 0,$$where $A^n = A \oplus \dotsc \oplus A$ as usual. It is not assumed that $A^n \to B^n$ is induced by some $A \to B$.



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> Assume that $C$ can be generated by $<n$ elements, does it follow $C=0$?
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I have proven this for $n=2$ [here](https://arxiv.org/abs/1205.2884) and numerical experiments suggest that it also holds for $n=3$. Of course the same question can be asked for finitely generated modules over a PID.



Exact sequence of $n$th powers of abelian groups
By Ycor's arguments, it seems we can reduce to $p$-groups.


Assume $A$ and $B$ are finite $p$-groups. The $p$-rank of $A$ must equal the $p$-rank of $B$ or else the right exact sequence $(C/pC) \to (A/pA)^n\to (B/pB)^n \to 0$ would imply that the $p$-rank of $C$ is at least $n$ and hence that $C$ has at least $n$ generators.


But given that the $p$-rank of $A$ equals the $p$-rank of $B$, the same right exact sequence implies that $(A/pA)^n\to (B/pB)^n$ is an isomorphism, and so the sequence $0 \to C \to (pA)^n \to (pB)^n \to 0$ exists and is exact. So we might as well work with $pA$ and $pB$.


We can now apply induction on the cardinality of $A$ and $B$ to reduce to the base case when $A=B=0$, which implies $C=0$.


Rather than using Ycor's reduction, one can instead rephrase the inductive argument as a proof that $(A / mA)^n \to (B / m B)^n$ is an isomorphism for all natural numbers $m$, which implies that $A^n \to B^n$ is an isomorphism by a standard lemma.




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Let me try to write carefully the argument for Noetherian rings. Let $C \to A^n \to B^n$ be an exact sequence of $R$-modules for a Noetherian ring $R$ where $C$ has fewer than $n$ generators. Our goal will be to show that $C$ vanishes. Let $\mathfrak m$ be a maximal ideal of $R$.


Consider the right exact sequence $(C/\mathfrak m C) \to (A/\mathfrak m A)^n\to (B/\mathfrak m B)^n \to 0$. We have $n \dim (A/\mathfrak m A) \leq \dim (C/\mathfrak m C) + n \dim(B/\mathfrak m B)$ and because every dimension is an integer and $\dim (C/\mathfrak m C) $, this implies that $\dim (A/\mathfrak m A) = \dim (B/\mathfrak m B)$ and hence the right map is an isomorphism, so $0 \to C \to (\mathfrak m A)^n \to (\mathfrak mB)^n \to 0$ is an exact sequence. 


Iterating this argument, we see that for all $k$, $(A/\mathfrak m^k A)^n \to (B/\mathfrak m^k B)^n$ is an isomorphism. Hence the induced map from the $\mathfrak m$-adic completion of $A$ to the $\mathfrak m$-adic completion of $B$ is an isomorphism. Because $\mathfrak m$-adic completion of a Noetherian local ring is an [exact functor](https://en.wikipedia.org/wiki/Completion_(algebra)#Properties) on finitely generated modules, this implies the $\mathfrak m$-adic completion of $C$ is zero. In particular, $C/\mathfrak m C$ is zero. Because this holds for all maximal ideals $\mathfrak m$, and $C$ is a finitely-generated module, $C$ vanishes.


