Let \(A\) be a finite abelian group of order \(n\). Prove that \(A\) is cyclic if and only if \(A\) has a unique subgroup of each order \(k\) such that \(k\) divides \(n\).
Solution
1. A cyclic group of order \(n\) has exactly one subgroupof order \(k\) for every positive divisor \(k\) of \(n\).
Write \(A=\langle g\rangle\). The element \(g^{n/k}\) has order \(k\): its \(j\)th power is one exactly when \(n\mid jn/k\), equivalently \(k\mid j\). Thus the subgroup \(\langle g^{n/k}\rangle\) has the required order.
For uniqueness, let \(H\leq A\) and choose the least positive integer \(t\) with \(g^t\in H\). Such an integer exists since \(g^n=1\in H\). If \(g^j\in H\), divide \(j\) by \(t\) to write \(j=qt+r\) with \(0\leq r<t\). Then \(g^r=g^j(g^t)^{-q}\in H\), and minimality forces \(r=0\). Applying this to \(j=n\) gives \(t\mid n\), and applying it to every element of \(H\) gives \(H=\langle g^t\rangle\). This subgroup has order \(n/t\). If that order is \(k\), then \(t=n/k\), proving uniqueness.
2. Under the uniqueness hypothesis, the primarydecomposition of \(A\) has at most one cyclic factor for each prime.
The structure theorem for finite abelian groups expresses \(A\) as a product of cyclic groups of prime-power order [@DF04]. Suppose two factors belonging to the same prime \(p\) occur, say \(C_{p^a}\) and \(C_{p^b}\) with \(a,b\geq1\). Use additive notation on the product. In the first factor, the class of \(p^{a-1}\) has order \(p\), so the element with that coordinate and all other coordinates zero generates a subgroup of order \(p\). The element with coordinate \(p^{b-1}\) in the second factor and zero elsewhere generates another subgroup of order \(p\).
These two subgroups are distinct: their nonzero elements have nonzero coordinates in different factors. Their images under the decomposition isomorphism remain distinct subgroups of \(A\). Since \(p\mid n\), this contradicts the assumed uniqueness for the divisor \(k=p\).
3. The group \(A\) is cyclic.
By step 2 its primary decomposition has the form \(A\cong\prod_{p\mid n}C_{p^{a_p}}\). Choose a generator in each factor. The resulting tuple has order equal to the least common multiple of the orders of its coordinates: a power of the tuple is one exactly when that power is divisible by every coordinate order. Since these orders are pairwise coprime, their least common multiple is \(\prod_{p\mid n}p^{a_p}=n\). The tuple therefore generates the entire order-\(n\) product, and its image generates \(A\). For \(n=1\), the group is trivial and generated by its identity, and its unique subgroup is itself. This also satisfies both directions of the assertion.