Proof that a finite abelian group is cyclic iff its exponent equals its order

Proof

Let $G$ be a finite abelian group and let $\exp(G)$ be its exponent.

1. By this result, the order of each $x \in G$ divides $|G|$. So $x^{|G|} = e$ for each $x \in G$, and hence $\exp(G) \leq |G|$.
2. Suppose that $G$ is cyclic, generated by $g$. By this result, $g$ has order $|G|$.
3. Since $g^{\exp(G)} = e$, the order $|G|$ of $g$ is at most $\exp(G)$. With step 1, $\exp(G) = |G|$.
4. Conversely, suppose that $\exp(G) = |G|$. If $G = \{e\}$, then $G$ is cyclic, generated by $e$. So let $G \neq \{e\}$.
5. By the classification of finite abelian groups, there is an isomorphism $G \cong A$ with $A = \Z/k_1\Z \oplus \cdots \oplus \Z/k_u\Z$, where $u \geq 1$ and $1 < k_1 \mid k_2 \mid \cdots \mid k_u$. Here $\oplus$ is the direct sum.
6. Isomorphic groups have the same order and the same exponent, since an isomorphism preserves products and the identity. So $\exp(A) = |A| = k_1 \cdots k_u$.
7. For each $x = (x_1, \dots, x_u) \in A$, we have $k_u x_i = 0$ in $\Z/k_i\Z$, because $k_i \mid k_u$. So $k_u x = 0$ and $\exp(A) \leq k_u$.
8. Suppose that $u \geq 2$. Then $|A| = k_1 \cdots k_u \geq k_1 k_u \geq 2 k_u > k_u \geq \exp(A)$. This contradicts step 6.
9. So $u = 1$ and $A = \Z/k_1\Z$, which is cyclic, generated by the class of $1$.
10. The image of a generator of $A$ under the inverse isomorphism generates $G$. So $G$ is cyclic.