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  "body": "\n## Proof\n\nLet $G$ be a finite abelian group and let $\\exp(G)$ be its [[exponent|0FCCTFA275]].\n\n1. By [[this result|08V7JHS3QZ]], the [[order|0JPYSD5E0B]] of each $x \\in G$ divides $|G|$. So $x^{|G|} = e$ for each $x \\in G$, and hence $\\exp(G) \\leq |G|$.\n2. Suppose that $G$ is [[cyclic|0B85Q41RP6]], generated by $g$. By [[this result|031M8HX7DA]], $g$ has order $|G|$.\n3. Since $g^{\\exp(G)} = e$, the order $|G|$ of $g$ is at most $\\exp(G)$. With step 1, $\\exp(G) = |G|$.\n4. Conversely, suppose that $\\exp(G) = |G|$. If $G = \\{e\\}$, then $G$ is cyclic, generated by $e$. So let $G \\neq \\{e\\}$.\n5. By the [[classification of finite abelian groups|07CQR9BW3R]], there is an [[isomorphism|085PND4KNT]] $G \\cong A$ with $A = \\Z/k_1\\Z \\oplus \\cdots \\oplus \\Z/k_u\\Z$, where $u \\geq 1$ and $1 \u003c k_1 \\mid k_2 \\mid \\cdots \\mid k_u$. Here $\\oplus$ is the [[direct sum|02WZB422CE]].\n6. Isomorphic groups have the same order and the same exponent, since an isomorphism preserves products and the identity. So $\\exp(A) = |A| = k_1 \\cdots k_u$.\n7. 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 $\\exp(A) \\leq k_u$.\n8. Suppose that $u \\geq 2$. Then $|A| = k_1 \\cdots k_u \\geq k_1 k_u \\geq 2 k_u \u003e k_u \\geq \\exp(A)$. This contradicts step 6.\n9. So $u = 1$ and $A = \\Z/k_1\\Z$, which is cyclic, generated by the class of $1$.\n10. The image of a generator of $A$ under the inverse isomorphism generates $G$. So $G$ is cyclic.\n\n## Idea\n\nA generator of a cyclic group has order equal to the group order, which forces the exponent up to the order. Conversely, in the invariant factor form the exponent is at most the largest factor $k_u$. The order is the product of all the factors, so the two can only agree when there is a single factor.\n",
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