Lemma 9.17.1. Let $A$ be an abelian group of exponent dividing $n$ such that $\{ x \in A \mid dx = 0\} $ has cardinality at most $d$ for all $d | n$. Then $A$ is cyclic of order dividing $n$.

**Proof.**
The conditions imply that $|A| \leq n$, in particular $A$ is finite. The structure of finite abelian groups shows that $A = \mathbf{Z}/e_1\mathbf{Z} \oplus \ldots \oplus \mathbf{Z}/e_ r\mathbf{Z}$ for some integers $1 < e_1 | e_2 | \ldots | e_ r$. This would imply that $\{ x \in A \mid e_1 x = 0\} $ has cardinality $e_1^ r$. Hence $r = 1$.
$\square$

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