Subgroups of groups of Square-free order - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:29:18Z http://mathoverflow.net/feeds/question/57217 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57217/subgroups-of-groups-of-square-free-order Subgroups of groups of Square-free order Martin David 2011-03-03T07:23:14Z 2011-03-03T10:43:49Z <p>If $G$ is a group of square-free order with at-least three prime factors, $|G|=p_1p_2....p_r$, $(2&lt; p_i &lt; p_{i+1})$, does $G$ contain a <strong>cyclic subgroup</strong> of composite order?</p> <p>(As groups of square-free order are solvable, $G$ will necessarily have a proper subgroup of composite order.)</p> http://mathoverflow.net/questions/57217/subgroups-of-groups-of-square-free-order/57223#57223 Answer by Tom De Medts for Subgroups of groups of Square-free order Tom De Medts 2011-03-03T08:50:24Z 2011-03-03T08:50:24Z <p>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).</p> <p>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.</p> http://mathoverflow.net/questions/57217/subgroups-of-groups-of-square-free-order/57238#57238 Answer by Guillermo Mantilla for Subgroups of groups of Square-free order Guillermo Mantilla 2011-03-03T10:43:49Z 2011-03-03T10:43:49Z <p>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: <strong>Fact:</strong> 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 $m$.</p> <p>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.</p>