$mA = 0 = nC, \ \gcd(m,n) = 1 \Rightarrow $ every extension of $A$ by $C$ splits

Question

This is Exercise 7.14(ii) from Rotman, Introduction to homological algebra, and I'm stuck on it.

If $A$ and $C$ are abelian groups, with $mA = 0 = nC $ and $\gcd(m,n) = 1$ then every extension of $A$ by $C$ splits, i.e., if we have the exact sequence $$0 \to A \to E \to C \to 0$$ then $E \cong A \oplus C.$

My thoughts: perhaps I should use the bijection between Ext$^1_{\mathbb{Z}}(C,A)$ and the extensions of $A$ by $C$. Moreover we have that $nA =A $ and $mC = C.$

Any hint ?

Answer 1

Here is another solution, using Ext groups. We only have to use that they are additive in each variable and that they classify extensions.

Since $m : A \to A$ vanishes, the same is true for $m : \mathrm{Ext}^1(C,A) \to \mathrm{Ext}^1(C,A)$. Likewise, $n$ vanishes on $\mathrm{Ext}^1(C,A)$. Since $n,m$ generate the unit ideal, it follows that $1$ vanishes on $\mathrm{Ext}^1(C,A)$. This means $\mathrm{Ext}^1(C,A)=0$.

Answer 2

It's more elementary than that. By Bézout's theorem, $\gcd(m,n) = 1 \implies \exists u,v : um+nv = 1$. Let $f : E \to E$ be defined by $f(x) = vnx$. Then since $nC = 0$, $$\operatorname{im}(f) \subset \operatorname{ker}(E \to C) = A$$ so $f$ is actually a morphism $E \to A$. And if $a \in A$, $$f(a) = vna = vna + u \underbrace{ma}_{=0} = (vn+um)a = a$$ So $f$ is a splitting of the exact sequence.