A Lemma in Serre's Arithmetic

Question

In Serre's A Course in Arithmetic, he writes the following lemma:

Lemma: Let $0 \rightarrow A \rightarrow E\rightarrow B\rightarrow 0$ be an exact sequence of commutative groups with $A$ and $B$ finite with orders $a$ and $b$ prime to each other. Let $B'$ be the set of $x\in E$ such that $bx=0$. The group $E$ is the direct sum of $A$ and $B'$.

Then in the proof, he says "since $bB'=0$, we have $bE\subset A$". I don't quite follow where this statement comes from. Can someone offer a hint?

Answer 1

Let $p:E\rightarrow B$ the projection. Remark that for every $x\in E$, whe have $p(bx)=0$ since the order of $B$ is $B$. This implies that $bx\in A$ since the sequence is exact, we deduce that a $abx =0$.

We have $1=ua+vb$ implies that for every $x\in E$, $x=uax+vbx$. We have $vbx=bvx\in A$, and $b(uax)=u(abx)=0$ im plies that $uax\in B'$.

Answer 2

HINT. Use exactness of the sequence.