A finite abelian group is cyclic iff its exponent equals its order

Statement

Let $G$ be a finite abelian group. Then $G$ is cyclic if and only if the exponent of $G$ equals $|G|$.

Remarks

If $G$ is cyclic with generator $g$, then the order of $g$ is $|G|$, so the exponent is $|G|$. For the converse, write $G \cong \mathbb{Z}/k_1\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/k_u\mathbb{Z}$ with $1 < k_1 \mid k_2 \mid \cdots \mid k_u$, by the classification. The exponent is $k_u$ and the order is $k_1 \cdots k_u$, so they are equal only when $u \leq 1$. The abelian condition is necessary: $S_3$ has exponent $6$ and order $6$, but it is not cyclic. For example, $(\mathbb{Z}/n\mathbb{Z})^\times$ is cyclic if and only if $\lambda(n) = \varphi(n)$.

Explanations

L1 human

In a cyclic group, one element has order equal to the size of the group. For a finite abelian group, it is enough that the exponent equals the size. Then the group is cyclic.