Direct sum of module and isomorphism Would you please help me to prove the following result?
If $A$ is an abelian group and $n\in \mathbb Z$ then we define
$$A[n]=\left\{a\in A| na=0\right\}$$
$$T(A)=\cup_{n\ge 1} A[n]$$
and if $p$ is a prime number $$T_p(A)=\cup_{k\ge 1} A[p^k]$$

Show that the map $x\mapsto (x_p)$ where $x_p$ is the $p-$th part of $x$ is an isomorphism from $T(A)$ to $\oplus_{p} T_p(A)$

( $x=x_px_{p′}$ where $x_p$ is the $p−$th part (has order a power of $p$) 
Thanks
 A: The existence of a $p$-th part (in some way) already assumes the correctness of the statement you are asking for, at least for finitely generated abelian groups. Let me try to give a more natural discussion of the underlying question.
There is a natural morphism (for example via definition of the direct sum) $$\bigoplus_p T_p(A) \rightarrow A$$ and one readily checks (for example by direct computation) that this factorizes as $$\bigoplus_p T_p(A) \stackrel{f}\rightarrow T(A) \hookrightarrow A.$$ 
I claim that $f$ is an isomorphism and we will actually use this isomorphism to define the $p$-th part of an element in $T(A)$ and it will then be clear that it's inverse is the morphism in question.
One can certainly formulate the following argument more elegantly; I will try to keep the notation as elementary as possible though.
Injectivity follows from the definition ($f$ is a direct sum of inclusions). So we need to show surjectivity. Let $x \in T(A)$ with $s \cdot x = 0$. If $s$ is of the form $p^n$, then $x \in T_p(A)$ and we are done. Now we do induction over the number of different primes appearing in the prime decomposition of $s$. Assume $s= p_1^{n_1} \cdots p_m^{n_m}$ and the statement holds true for upto $m-1$ factors for elements $x$ in arbitry abelian groups. By induction, we find $y' \in f^{-1}(<x>)$ such that $f(y')=p_1^{n_1}x$. On the other hand, again by induction, we find $y'' \in f^{-1}(<x>/<p_1^{n_1}x>)$ such that $f(y'')= x + c \cdot p_1^{n_1} x$ for some $c \in \mathbb{Z}$. Now $y'' - cy'$ yields the desired preimage.
