This is not a homework!

I am trying to solve the following exercise of the book "Permutation Groups" by Dixon and Mortimer:

enter image description here

I know that $p \mid \vert G\vert$, because $G$ is transitive of degree $p$. So, there is an element $x\in G$ with $o(x)=p$. Therefore, we can suppose that $R=\langle (1,2,\ldots,p)\rangle$ is a supgroup of $G$. It is simple to prove that $R$ is a regular permutation subgroup of $G$.

Using these facts, I am trying to show that if $G$ is solvable, $R$ is a normal Sylow $p$-subgroup of $G$, and since $p \nmid \vert \frac{G}{R} \vert$, $G$ can be consider as a semi-direct product group. I don't know that this direction is useful for solving this exercise or not.

I will be so grateful for your helpful guidance and answers.

  • $\begingroup$ What does 'degree' mean in this case? The order of the set being acted on by $G$? $\endgroup$ – john Dec 28 '15 at 0:12
  • 1
    $\begingroup$ Assume that $G$ is solvable. Then it has a nontrivial normal abelian subgroup $N$. There is a theorem (which you should know) that the orbits of a normal subgroup of a transitive permutation group all have the same size and, since the degree $p$ is prime, this size must be $p$. So $N$ is transitive and, since it is abelian, it must be regular. That gives you (ii). $\endgroup$ – Derek Holt Dec 28 '15 at 10:17
  • $\begingroup$ @john: Yes. That is right. $\endgroup$ – A-213 Dec 28 '15 at 12:33
  • $\begingroup$ @DerekHolt: Nice answer! Thank you very much. I think that by having (ii), we can say that $G=N\rtimes G_{\alpha}$, where $\alpha\in \Omega$. Also, I know that $AGL_1(p)\cong Z_p\rtimes Z_p^*$, but I don't know how I can make the desired isomorphism. I will be so grateful if you guide me about this part, too. $\endgroup$ – A-213 Dec 28 '15 at 12:54
  • 1
    $\begingroup$ ${\rm AGL}(1,p)$ is the normalizer in $S_p$ of a Sylow $p$-subgroup of $S_p$ (just compare their orders), so (ii) implies (iii). $\endgroup$ – Derek Holt Dec 28 '15 at 13:31

Your Answer

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

Browse other questions tagged or ask your own question.