## Characterization of the natural numbers which are only the orders of cyclic groups, part 2

Prove the converse to the preceding exercise: if $n \geq 2$ is an integer such that every group of order $n$ is cyclic, then $n = p_1p_2 \cdots p_k$ is a product of distinct primes $p_i$ such that $p_i$ does not divide $p_j - 1$ for all $i,j$. [Hint: If $n$ is not of this form, construct noncyclic groups using direct products and known noncyclic groups of order $p^2$ and $pq$ where $p|q-1$.]

Let $n = p_1p_2 \cdots p_k$, where the $p_i$ are prime.

• Suppose (without loss of generality) that $p_1 = p_2$. Then $Z_{p_1}^2 \times Z_{p_3 \cdots p_k}$ is noncyclic since (in particular) it has a noncyclic subgroup $Z_{p_1}^2$. Moreover, this group has order $n$. Thus not every group of order $n$ is cyclic.
• Suppose (without loss of generality) that $p_1$ divides $p_2 - 1$. By the example on page 143 in the text, there exists a nonabelian group $Q$ of order $p_1p_2$. Now $Q$ is not cyclic, since cyclic groups are abelian, hence $Q \times Z_{p_3 \cdots p_k}$ is a noncyclic group of order $n$. Thus not every group of order $n$ is cyclic.

Hence, if all groups of order $n$ are cyclic, then the $p_i$ are mutually distinct and $p_i$ does not divide $p_j-1$ for all $i,j$.