Timeline for Subgroups of groups of Square-free order
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 15, 2011 at 7:08 | vote | accept | Martin David | ||
| Mar 13, 2011 at 9:59 | vote | accept | Martin David | ||
| Mar 15, 2011 at 7:08 | |||||
| Mar 6, 2011 at 21:49 | comment | added | Tom De Medts | @Martin: This is just a way of rephrasing your original question in the case $r=3$; or am I missing something? | |
| Mar 5, 2011 at 8:29 | comment | added | Martin David | @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 $G$)? | |
| Mar 3, 2011 at 16:13 | comment | added | Steve D | 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. | |
| Mar 3, 2011 at 11:50 | comment | added | Tom De Medts | 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). | |
| Mar 3, 2011 at 11:16 | comment | added | Martin David | @ Tom: I would like to see the relations between $m,n,r$. Can you suggest reference for it, please? | |
| Mar 3, 2011 at 8:50 | history | answered | Tom De Medts | CC BY-SA 2.5 |