How do you prove that the torsion subgroup homomorphism $\bigoplus_p A(p) \rightarrow A$ is surjective.

How do you prove that the torsion subgroup homomorphism $\bigoplus_p A(p)
\rightarrow A$ is surjective.

Here's Lang's Algebra book's proof:
Theorem 8.1 Let $A$ be a torsion abelian group. Then $A$ is the direct sum
of its subgroups $A(p)$ for all primes $p$ such that $A(p) \neq 0$.
(Jump to the second paragraph)
Proof. There is a homomorphism $\bigoplus_p A(p) \rightarrow A$ which to
each element $(x_p)$ in the direct sum associates the element $\sum x_p$
in $A$. We prove that this homomorphism is both surjective and injective.
Suppose $x$ is in the kernel, so $\sum x_p = 0$. Let $q$ be a prime. Then
$ x_q = \sum_{p \neq q} (-x_p)$ Let $m$ be the least common multiple of
the periods of elements $x_p$ on the right- hand side, with $x_q \neq 0$
and $p \neq q$. Then $mx_q = 0$. But also $q^rx_q = 0$ for some positive
integer $r$. If $d$ is the greatest common divisor of $m$, $q_r$ then
$dx_q = 0$, but $d = 1$, so $x_q = 0$. Hence the kernel is trivial, and
the homomorphism is injective.
As for the surjectivity, for each positive integer $m$, denote by $A_m$
the kernel of multiplication by $m$, i.e. the subgroup of $x \in A$ such
that $mx = 0$. We prove: If $m = rs$ with $r, s$ positive relative prime
integers, then $A_m = A_r + A_s$. Indeed, there exist integers $u$, $v$
such that $ur + vs = 1$. Then $x = urx + vsx$, and $urx \in A_s$ while
$vsx \in A_r$ and our assertion is proved. Repeating this process
inductively, we conclude: If $m= \prod_{p|m} p^e(p)$, then $A_m =
\sum_{p|m} A_{p^{e(p)}}$. Hence the map $\bigoplus_p A(p)
\twoheadrightarrow A$ is surjective, and the theorem is proved.
Edit
I think he wants to say that since $A$ is torsion each $x\in A$ has finite
order $m$. But $x \in A_m = \sum_{p|m} A_{p^{e(p)}}$, and since each
$A_{p^{e(p)}} \leqslant A(p)$ we have that their exist representatives
$x_p \in A(p)$ such that $\sum x_p = x$. QED.