Exercise 4.5.54 (Classification of groups of order $p_1\cdots p_r$, where $p_i \nmid p_j - 1$)

Proof. If $|G|=p$, where $p$ is a prime, then $G$ is cyclic.

• Reasoning by induction, suppose that every group of order less than $n$, satisfies the statement. Let $G$ be a group of order $n=p_1{p}_2\cdots p_r(r>1)$, where the $p_i$’s are distinct primes such that $p_i$ does not divide $p_j-1$ for all $i$ and $j$. If $H$ is a proper subgroup of $G$, then $|H|$ divides $|G|$, thus $|H|=q_1{q}_2\cdots q_l<n$, where $\{q_1,q_2,\dots ,q_l\}\subset \{p_1,p_2,\dots ,p_r\}$, so that $q_1,q_2,\dots ,q_l$ are distinct primes such that $q_i$ does not divide $q_j-1$ for all $i$ and $j$. The induction hypothesis shows that $H$ is cyclic, so every proper subgroup of $G$ is cyclic.
• If $G$ is not abelian, $G$ is not simple by Exercise 53. If $G$ is abelian, let $a$ be an element of order $p_1$, which exists by Cauchy’s Theorem. Then $⟨a⟩$ is a non trivial proper (normal) subgroup. In both cases, $G$ is not simple. Therefore there exists some nontrivial proper normal subgroup $N$ of $G$. Then $N$ is cyclic.

• Since $N⊴G$, $G$ acts by conjugation on $N$. This action affords a homomorphism

  $$\psi :G\to \mathrm{Aut}(N).$$

  (Note that $N\le \ker <!--\; FUNCTION\; APPLICATION\; -->(\psi )$: since $N$ is abelian, every $g\in N$ acts trivially on $N$. Therefore $\psi$ induces a homomorphism $\lambda :G∕N\to \mathrm{Aut}(N)$ such that $\lambda \circ \pi =\psi$, where $\pi :G\to G∕N$ is the natural projection. We don’t use this argument.)

• Note that $|G|=p_1{p}_2\dots p_r$, and by Lagrange’s Theorem, $s=|N|=p_{i_1}\dots p_{i_l}$, where $i_1,i_2,\dots ,i_l\in [[1,r]]$ are distinct.

  By Proposition 16, since $N$ is cyclic,

  $$|\mathrm{Aut}(N)|\simeq {(\mathbb{Z}∕\mathit{𝑠\mathbb{Z}})}^{\times }.$$

  Therefore

  $$|\mathrm{Aut}(N)|=\phi (p_{i_1}\dots p_{i_r})=(p_{i_1}-1)(p_{i_2}-1)\cdots (p_{i_l}-1).$$

  For every indices $i,j$, $p_i\nmid p_{i_j}-1$ by hypothesis, therefore

  $$g.c.d.(G,|\mathrm{Aut}(N)|)=1.$$

  This implies that the homomorphism $\psi$ is trivial: $|\psi (G)|$ is a divisor of $|\mathrm{Aut}(N)|$ and of $|G|$, so $\psi (G)=\{{\mathrm{id}}_N\}$. Hence for all $g\in G$, $\phi (g)={\mathrm{id}}_N$, so that $\mathit{𝑔𝑛}{g}^{-1}=n$ for all $n\in N$. This proves

  $$N\le Z(G).$$

• Since $N$ is not trivial, $Z(G)$ is not trivial, thus $|G∕Z(G)|<n=|G|$, and $|G∕Z(G)|$ is a product of some $p_i,1\le i\le r$. The induction hypothesis shows that $G∕Z(G)$ is cyclic. By Exercise 3.1.36, $G$ is abelian.

• Since $G$ is abelian, and $|G|=p_1{p}_2\cdots p_r$, where the $p_i$’s are distinct primes, then

  $$G\simeq \mathbb{Z}∕p_1\mathbb{Z}\times \mathbb{Z}∕p_2\mathbb{Z}\times \cdots \times \mathbb{Z}∕p_r\mathbb{Z}\simeq \mathbb{Z}∕\mathit{𝑛\mathbb{Z}},$$

  so $G$ is cyclic (By Cauchy’s Theorem, there is some $a_i$ of order $p_i$, then $a_1{a}_2\cdots a_r$ has order $p_1{p}_2\cdots p_r=n$).

The induction is done, which proves that if $G$ is a finite group of order $p_1{p}_2\cdots p_r$ where the $p_i$’s are distinct primes such that $p_i$ does not divide $p_j-1$ for all $i$ and $j$, then $G$ is cyclic. □