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.)


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.

| cite | improve this answer | |
  • $\begingroup$ @ Tom: I would like to see the relations between $m,n,r$. Can you suggest reference for it, please? $\endgroup$ – Martin David Mar 3 '11 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 Mar 3 '11 at 11:50
  • 1
    $\begingroup$ 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 Mar 3 '11 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 Mar 5 '11 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 Mar 6 '11 at 21:49

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.

| cite | improve this answer | |
  • $\begingroup$ @ 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 Mar 3 '11 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 Mar 3 '11 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 Mar 3 '11 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 Mar 3 '11 at 11:53
  • $\begingroup$ @Tom: Who proved this result? (Ref.?) $\endgroup$ – Martin David Mar 3 '11 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.