On infinite abelian $p$-group of bounded order Definition. If $p$ is a prime, then a $p$-group is a group in which every element has order a power of $p$.
Remark: An additively writen group is called bounded if its elements have boundedly finite orders. Of course multiplicative groups with this property are said to have finite exponent but this term is inappropriate in the context of additive groups.
Let $G$ be an infinite abelian $p$-group of bounded order, then prove that $G\cong
\mathbb{Z}_{p^{n}}\oplus\mathbb{Z}_{p^{n}}\oplus H$, for some natural number $n$ and for some abelian group
$H$.
 A: The "bounded order" you mention is usually called the exponent of the group.  So you are asking about an infinite abelian p-group of bounded exponent.  Try using the following theorem of Prufer:
Theorem: If $A$ is an abelian p-group of bounded exponent, then $A$ is a direct sum of cyclic groups.
Sketch of Proof:  Let $p^n$ be the exponent, and induct on $n$.  The base case $n=1$ is a corollary of the fact $A$ is then a vector space over $\mathbb{Z}/p\mathbb{Z}$.  
For the inductive step, consider $pA$, which is a direct sum of cyclic groups.  Let $pA=\oplus A_i$, with the $A_i$ cyclic and generated by $a_i$.  Let $pb_i=a_i$.  Then show $B$ (generated by the $b_i$) is a direct sum of cyclic groups.
Finally, pick a subgroup $C$ of $A$ maximal with respect to satisfying $C\cap B=\lbrace0\rbrace$.  Show $A=B\oplus C$.
To get the statement you gave, write $A=\oplus X_i$ as the direct sum of cyclic subgroups.  Note that each $X_i\cong \mathbb{Z}/p^m\mathbb{Z}$ for some $m\le n$; if there was only (at most) one copy of each, $A$ would be finite.
