$|G|=p_1p_2p_3$ distinct primes with $p_i \nmid p_j-1$ then $G$ is cyclic

Question

Problem Let $p_1,p_2,p_3$ be three distinct primes with $p_i \nmid p_j-1$ for all $1\leq i,j \leq 3$ and let $G$ be a group of order $p_1p_2p_3$. Show that $G$ is cyclic.

I've tried to come up with a solution but I got stuck at one point. My idea was to prove that there is a unique $p_i$-Sylow subgroup for $i=1,2,3$. If I could show this, then $H,K,S$ the unique Sylow subgroups for each prime divisor of the order of $G$ are normal in $G$. But then $G \cong H \times K \times S \cong \mathbb Z_{p_1} \oplus \mathbb Z_{p_2} \oplus \mathbb Z_{p_3} \cong\mathbb Z_{p_1p_2p_3}$. From here it follows $G$ is cyclic.

Let $r_{p_i}=\{\text{number of $p_i$-Sylow subgroups}\}$. My goal is to prove $r_{p_i}=1$.

By the Sylow theorems, $r_{p_i} \equiv 1 (p_i)$ and $r_{p_i} \mid p_jp_k$ $\implies r_{p_i} \in \{1,p_j,p_k,p_jp_k\}$. Since $p_i \nmid p_j-1$, then $r_{p_i} \in \{1,p_jp_k\}$. Suppose there is some $i \in \{1,2,3\}$ : $r_i=pjpk$, without loss of generality $r_{p_1}=p_2p_3$, then $$p_1p_2p_3=|G| \geq r_{p_1}(p_1-1)+r_{p_2}(p_2-1)+r_{p_3}(p_3-1)$$ $$\geq p_1p_2p_3-p_2p_3+p_2-1+p_3-1+1$$$$=p_1p_2p_3-p_2p_3+p_2-1+p_3$$

I don't know how to arrive to a contradiction, I would appreciate suggestions to complete my idea or maybe another approach to show $r_{p_i}=1$ for all $i=1,2,3$. Thanks in advance.

Answer 1

I'm not sure restricting to three prime factors makes this classical (but nontrivial) result really easier.

See here for a proof of the general result.

Answer 2

Let Np1 be the number of Sylow p1-subgroups of G. Np1 must be 1mod p1 and it must divide m=p2*p3. This can happen with Np1 unequal to 1, for example, p1=3, p2=5, p3=17 would allow Np1=N3=85. So a group G of order 3*5*17 need NOT be abelian, although 3 does not divide 5-1 or 17-1 and 5 does not divide 17-1. But 3 does divide 5*17-1. Hope that helps.