IV, 5.1, footnote, simpler proof Let $A$ be an abelian group such that, for some prime $p$, (a) $A$ is $p$-primary (every element is killed by a power of $p$), (b) there are only finitely many elements of order $p$, (c) $\bigcap_n p^n A=0$. Then $A$ is finite. Because of (a) and (b), it suffices to show that $A$ is bounded, i.e., that $p^nA=0$ for some $n$. If not, there exists an infinite sequence of elements $a_1,a_2,...$ such that $a_i$ has order $p^i$. But then the elements $p^{i-1}a_i$ are each of order $p$ and so, by (b), at least one of them must occur for infinitely many $i$, contradicting (c). (sx4939604, Siefert).