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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.

Community
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user156441
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    We typically say "arrive at a contradiction" rather than "arrive to an absurd". Sorry that doesn't answer your question, but just thought you might want to know. – Gregory Grant May 05 '15 at 14:29
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    If you start using Sylow for the smallest of the three primes, say $p_1$, you either get a normal $p_1$-Sylow subgroup (and finish off by factoring it out) or so many (how many?) elements of order $p_1$ that there aren't enough for more than one $p_2$- or $p_3$-Sylow group (and finish off by showing that elements of order $p_1$ have to centralize normal $p_2$- or $p_3$-Sylow groups). – j.p. May 05 '15 at 15:39

2 Answers2

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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.

PseudoNeo
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  • That's an interesting induction argument, not on the number of primes but on the actual product of distinct primes. It seems there must be an easier way to show $G$ is abelian, if it had two elements that don't commute then by 3 they must generate the entire group. From there it just seems like there should be an easier way, but I didn't think about it too long. – Gregory Grant May 05 '15 at 14:37
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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.

Richard Peterson
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  • Or, maybe I made a mistake in my argument above! The reference PseudoNeo gave says it must be cyclic. – Richard Peterson May 06 '15 at 02:02
  • Sylow's theorems say that N3 in the case above must be one of {1, 85}. Sylow does not say it CAN take on both of those values, as far as I know. Maybe that was my mistake, if it turns out that G must be abelian. – Richard Peterson May 06 '15 at 02:11