I'm trying to understand this fact, but I have problem with a part of the proof.
Let $0 \to A \to E \to B \to 0$ be an exact sequence of commutative groups (denoted additively) with $A$ and $B$ finite of orders $a$ and $b$ prime to each other. Let $B':= \{x \in E|bx = 0\}$. Then $E$ is the direct sum of $A$ and $B'$.
The proof begins proving that $A \cap B'=\{0\}$. Then, for every $x \in E$, I can write $x=(ar+bs)x$ with $r,s \in \mathbb{Z}$ using Bézout. Now the book says that, since $bB'=\{0\}$, $bE \subseteq A$, but I don't understand why. Can someone help me? Thanks!
One way to see that $bE\subseteq A$ is to note that $bE\subseteq\ker(E\ \rightarrow\ B)$ because $b=|B|$ annihilates $B$. Because the sequence is exact we get $bE\subseteq\ker(E\ \rightarrow\ B)=\operatorname{im}(A\ \rightarrow\ E)$, i.e. $bE\subseteq A$.
How this relates to the fact that $bB'=\{0\}$ I do not see at the moment.