Let $A$ be an abelian group. $A(p)$ is a p-group if $A(p)$ is finite. Let $p$ be a prime number, let $A$ be an abelian group, define $A(p)$ to be the subgroup of all elements that have power of $p$ order, i.e. $x \in A(p) \iff x\in A \ \wedge \ p^ex = 0$, in additive group notation.  Note that for some $e$, $p^ex = 0 \iff p^{ord(x)}x = 0$, so that statement is correct.  Now I want to prove that if $A(p)$ is finite then it's a $p$-group in the most general setting, so I was thinking: if $A$ is any finte group then there is an element of order $|A|$, but that's not true since not every finite group is cyclic.  I have no idea what makes $A(p)$ special so that it's a $p$-group when finite.
For each $x \in A(p)$, $ \ x \mathbb{Z}$ is a finite cyclic of order $p^e$ where $e = ord(x)$ in $A(p)$.  We can write $A(p) = x_1\mathbb{Z} + \dots + x_n\mathbb{Z}$, where $x_i$ runs through the nonzero elements of $A(p)$.  Let's try to create a basis.  If $k_1 x_1 + \dots + k_n x_n = 0$, then $k_n x_n = $ sum of rest of terms.   If $x_n$ has order $p^e$, then without loss of generality, let $k_n \lt p^e$, since if not, then $k_n$ can be reduced to that since $(mp^e + k_n')x_n = 0 + k_n' x_n$ where $k_n'\lt p^e$.  Almost there, still working on it...
Edit
Thanks, TBrendle.  There are two definitions of $p$-group and they're equivalent.  A finite group is a $p$-group iff all of its elements have $p$-power order.  One direction is obvious. To show the converse, if $|A| = qm$ for any prime $q$ other than $p$ then it has an element of order $q$, by another theorem.  Thus $|A| = p^e$ for some $e$.

A $p$-group is usually defined as a group in which all elements have $p$-power order, so this is true by definition.
If you are working with an alternative definition where $p$-group means a finite group whose order is a power of $p$: If $A(p)$ is not a $p$-group then by Cauchy's theorem there is some element in $A(p)$ whose order is a prime $q$ with $q \neq p$, but this is a contradiction to the definition of $A(p)$.
If you are having trouble with this problem, you may not understand the definitions involved.
