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If $G$ is a group of square-free order with at-least three prime factors, $|G|=p_1p_2....p_r$, $(2< p_i < p_{i+1})$, does $G$ contain a cyclic subgroup of composite order?

(As groups of square-free order are solvable, $G$ will necessarily have a proper subgroup of composite order.)

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Yes, $G$ always contains a cyclic subgroup of composite order. Note that all Sylow subgroups of $G$ are cyclic, i.e. $G$ is a Zassenhaus metacyclic group. Such groups have a very precise structure: they are of the form $$ G = \left\langle a, b \mid a^m = b^n = 1, b^{-1} a b = a^r \right\rangle ,$$ where $m,n,r$ satisfy certain restrictions that are not important now. Such a group has order $G = mn$ (in fact more can be said: the derived subgroup $G'$ has order $m$, and both $G'$ and $G/G'$ are cyclic).

Since at least one of the numbers $m$ or $n$ is composite, it follows that either $\langle a \rangle$ or $\langle b \rangle$ is a cyclic subgroup of $G$ of composite order.

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  • $\begingroup$ @ Tom: I would like to see the relations between $m,n,r$. Can you suggest reference for it, please? $\endgroup$
    – Martin David
    Commented Mar 3, 2011 at 11:16
  • $\begingroup$ The relations are: $\gcd(m,n) = 1$, $\gcd(m, r-1) = 1$ and $r^n \equiv 1 \pmod{m}$. A possible reference is Huppert's "Endliche Gruppen I", Chapter IV, Satz 2.11 (p. 420). $\endgroup$
    – Tom De Medts
    Commented Mar 3, 2011 at 11:50
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    $\begingroup$ I don't think it's necessary to use this result for this simple case: It can be shown the Sylow $p_r$-subgroup $P$ is normal, and there is a complement $M$ in $G$; that is, $G=P\rtimes M$, and either (i) $C_M(P)= 1$, which means $M$ embeds in $Aut(P)$ (which is cyclic), or (ii) $1\neq x\in C_M(P)$ has prime order, so $\langle x,P\rangle$ is cyclic. $\endgroup$
    – Steve D
    Commented Mar 3, 2011 at 16:13
  • $\begingroup$ @Tom : I want to make one step clear towards the Answer: Suppose we have a group $G$ of order $pqr$, where $p<q<r$ are odd primes such that $p|(q-1)$, $p|(r-1)$, and $q|(r-1)$. As a group of this order is solvable, so by Halls theorem, $G$ has subgroups of order $pq$, $pr$, and $qr$, and subgroup of same order are conjugate. But as we have chosen primes $p,q,r$ with the above relations, can it happen that subgroup of order $pq$ is $C_q \rtimes C_p$, subgroup of order $pr$ is $C_r \rtimes C_p$, and subgroup of order $qr$ is $C_r\rtimes C_q$ (simultaneously in $G$)? $\endgroup$
    – Martin David
    Commented Mar 5, 2011 at 8:29
  • $\begingroup$ @Martin: This is just a way of rephrasing your original question in the case $r=3$; or am I missing something? $\endgroup$
    – Tom De Medts
    Commented Mar 6, 2011 at 21:49
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Adding a bit to Tom's very complet answer you can in fact find a cyclic subgroup of oder $pq$ where $p,q$ are among the three bigest primes in the factorization of $|G|$. First notice that $G$ contains a subgroup $H$ of order $p_{r-2}p_{r-1}p_r$. This follows inductively from: Fact: If $|G|=pm$ where $(p,m)=1$ and $p$ is the smallest prime dividing $|G|$ then $G$ contains a subgroup, in fact normal, of order $m$.

Now you've got a group $H$ of order $p_{r-2}p_{r-1}p_r$, so you use Tom's argument with $H$ and obtain that $m$ or $n$ is $pq$. I believe that for groups of order $pqr$ one could actually show by hand what you want without invoking the known presentation of Z-groups.

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  • $\begingroup$ @ Guillermo: A small doubt about FACT : if |G|=pm, where (p,m)=1, and p is the smallest prime dividing order of |G|, then a subgroup of G of index p (i.e. of order m), if exists, must be normal; but how can we say that such a subgroup exists? – Martin David 0 secs ago $\endgroup$
    – Martin David
    Commented Mar 3, 2011 at 11:13
  • $\begingroup$ It is a result of Hall that a finite group is solvable if and only every p-sylow subgroup has a complement(I've seen this result in one of Isaacs' books, either the one of graduate algebra or finite group theory). If $|G|=pm$ where $p$ and $m$ are as above then $G$ is solvable. If $p=2$ it is easy to get a complement, which is obviously normal so $G$ is an extension of a $2$-group by an odd order group so $G$ is solvable. If $p$ is odd then the group is an odd order group so it is solvable. Now since $G$ is solvable, and a p-Sylow has oder $p$ it must have a complement. I'm almost sure... $\endgroup$
    – Guillermo Mantilla
    Commented Mar 3, 2011 at 11:30
  • $\begingroup$ ...there is an easier way to construct a subgroup of index $p$ -in a similar way to what one would do for $p=2$(I might be wrong). Anyway, if there is another way to do it you'll find it for sure in Isaacs(Finite group theory) $\endgroup$
    – Guillermo Mantilla
    Commented Mar 3, 2011 at 11:35
  • $\begingroup$ In fact, any group $G$ for which all Sylow subgroups are cyclic, contains a subgroup of order $m$ for any divisor $m$ of $|G|$. $\endgroup$
    – Tom De Medts
    Commented Mar 3, 2011 at 11:53
  • $\begingroup$ @Tom: Who proved this result? (Ref.?) $\endgroup$
    – Martin David
    Commented Mar 3, 2011 at 11:59

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