Timeline for Subgroups of groups of Square-free order
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Jun 27, 2016 at 15:39 | comment | added | yakov | Our group $G$ is supersolvable. So, by Wendt's theorem, $G'$ is nilpotent so cyclic. In that case, at least one of groups $G'$, $G/G'$ is of composite order. This is also true provided all Sylow subgroups of $G$ are cyclic and $|G|$ is a product of $>2$ primes. | |
| Mar 15, 2011 at 7:08 | vote | accept | Martin David | ||
| Mar 15, 2011 at 7:08 | |||||
| Mar 3, 2011 at 15:16 | comment | added | Jack Schmidt | This stuff is covered in several group theory textbooks. Check M. Hall, Jr.'s Theory of Groups, especially 9.4, 10.5, and 14. A group of order pm for (p,m)=1 and p the smallest prime divisor is p-nilpotent by Burnside, Frobenius, and co. For p=2 this is Cayley. The subgroup of order m is the kernel of the transfer into a Sylow p-subgroup. This is in both Robinson's and Isaacs's group theory textbooks as well, I believe. | |
| Mar 3, 2011 at 12:56 | comment | added | Tom De Medts | The fact that every such group has the structure of the form that I've explained in my answer, is due to Zassenhaus (hence the name Zassenhaus metacyclic groups or $Z$-groups for short). The statement that for every divisor of $|G|$ there is a corresponding subgroup, follows easily from this description (write this divisor as $m'n'$ with $m' \mid m$ and $n' \mid n$, and take $r' = r^{n / n'}$). Alternatively, in your specific case you can simply invoke Hall's theorem saying that a solvable group $G$ has a Hall $\pi$-subgroup for every set $\pi$ of primes. | |
| Mar 3, 2011 at 11:59 | comment | added | Martin David | @Tom: Who proved this result? (Ref.?) | |
| Mar 3, 2011 at 11:53 | comment | added | Tom De Medts | In fact, any group $G$ for which all Sylow subgroups are cyclic, contains a subgroup of order $m$ for any divisor $m$ of $|G|$. | |
| Mar 3, 2011 at 11:35 | comment | added | Guillermo Mantilla | ...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) | |
| Mar 3, 2011 at 11:30 | comment | added | Guillermo Mantilla | 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... | |
| Mar 3, 2011 at 11:13 | comment | added | Martin David | @ 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 | |
| Mar 3, 2011 at 10:43 | history | answered | Guillermo Mantilla | CC BY-SA 2.5 |