Exercise 4.5.55 (Converse to the preceding exercise)

Proof. Let $G$ be a group of order $n\ge 2$.

• If ${\nu }_p(n)>1$ for some $p$, then $|G|=p^{2}u$, for some integer $u$. Consider the group

  $$\begin{array}{lll}\hfill G=Z_p\times Z_p\times Z_u.& & \hfill \end{array}$$

  For all $x\in G_2$, $x^{\mathit{𝑝𝑢}}=1$, thus there is no element of order $n=p^{2}u$ in $G$, so $G$ is not cyclic.

• Suppose that there are two prime divisors $p,q$ of $n=|G|$ such that $p\mid q-1$. Then there exists a non abelian group $H$ of order $\mathit{𝑝𝑞}$ (see the proof in “Example, groups of order $\mathit{𝑝𝑞}$” p. 143, or in Chapter 5, where $H$ is constructed as a semi-direct product).

  Then the group

  $$G^{\prime }=H\times Z_v$$

  has order $n=\mathit{𝑝𝑞𝑣}$. Moreover $G^{\prime }$ is not abelian, so $G^{\prime }$ is not cyclic.

This proves that if every group of order $n$ is cyclic, then $n=p_1{p}_2\cdots p_r$ is a product of distinct primes and $p_i$ does not divide $p_j-1$ for all $i,j$. □