Let \(G\) be a finite abelian group. Suppose that for every positive integer \(n\), at most \(n\) elements of \(G\) satisfy \(x^n=e\) (equivalently, have order dividing \(n\)). Prove that \(G\) is cyclic.
Solution
Let \(m\) be the exponent of \(G\), the least positive integer such that \[ x^m=e\qquad\text{for every }x\in G. \] Since every element of \(G\) satisfies \(x^m=e\), the hypothesis with \(n=m\) gives \[ |G|\le m. \] On the other hand, the order of every element divides \(|G|\) by Lagrange’s theorem, so the exponent also divides \(|G|\). Hence \[ m\le |G|. \] Therefore \[ m=|G|. \]
For a finite abelian group, the exponent is attained as the order of some element. Indeed, decompose \(G\) into its Sylow subgroups \[ G=\prod_p G_p. \] For each \(p\), choose an element \(g_p\in G_p\) of maximal order, equal to the exponent of \(G_p\). Since these orders are pairwise coprime, the element \[ g=\prod_p g_p \] has order equal to the product of those exponents, namely \(m\).
Thus \(G\) contains an element of order \[ m=|G|. \] That element generates all of \(G\), so \(G\) is cyclic.