If the order of a finite abelian group is not divisible by a square, show that the group must be cyclic. If the order of a finite abelian group is square free, show that the group is cyclic.
This is a question from "basic abstract algebra" by bhattacharya
 A: $|G| = p_1 \dots p_n$, a product of distinct primes. Use Cauchy's theorem to get $g_i \in G$ of order $p_i$ for $1 \leq i \leq n$. Then, use induction and the assumption that $G$ is abelian to prove $g_1\ldots g_i$ has order $p_1 \ldots p_i$.
A: The structure theorem for finitely generated abelian groups says that if $G$ is finitely generated and abelian, then it is a product of finitely many groups of the form $\mathbb{Z}$ and $\mathbb{Z}_q = \mathbb{Z}/q\mathbb{Z} $ where $q$ is power of a prime. If $G$ is finite, then surely it is finitely generated and has no factor isomorphic to $\mathbb{Z}$; thus it is of the form:
$$ G \simeq \mathbb{Z}_{q_1} \times \mathbb{Z}_{q_2} \times \dots \times \mathbb{Z}_{q_n} $$ 
where $q_i$ are (not necessarily distinct) powers of primes. It is clear that $|G| = \prod_{i=1}^n q_i$. Because $|G|$ is square-free, $q_i$ have to be all primes (i.e. not some higher powers) and distinct. In particular, $q_i$ and $q_j$ are compime for $i \neq j$. The Chinese Remainder Theorem now yields:
$$ \mathbb{Z}_{q_1} \times \mathbb{Z}_{q_2} \times \dots \times \mathbb{Z}_{q_n} \simeq \mathbb{Z}_{q_1 q_2 \dots q_n} $$
This is a cyclic group.
