I noticed that the following is true, and that there is a reasonably elementary proof of it (in particular, the classification of finite simple groups is not needed). Let $G$ be a finite permutation group which contains two $p$-cycles which do not commute (where $p$ is any odd prime other than a Mersenne prime). Then $G$ is not solvable ( more precisely, $G$ has a non-Abelian composition factor of order divisible by $p$). Since the methods are reasonably elementary, I wonder if anyone has come across this or similar results (possibly in a Galois Theory context) and can point me to a reference?

For every Mersenne prime $p,$ there is a solvable permutation group $G$ of degree $p+1$ and order $p(p+1)$ which contains two non-commuting $p$-cycles.

(Later note: The analogous result is not true for $p^{2}$-cycles ($p$ prime). For $p=2$ take $G = S_{4}$ and for $p>2$ take $G$ to be a Sylow $p$-subgroup of $S_{p^{2}}.$ In each case, $G$ is a solvable (even nilpotent when $p$ is odd) permutation group containing two $p^{2}$-cycles which do not commute).

  • $\begingroup$ So you say e.g. that ${\rm A}_4$ is solvable because $3$ is a Mersenne prime? $\endgroup$ – Stefan Kohl Jun 18 '14 at 15:36
  • $\begingroup$ Well, I don't say that, but the smallest exceptional case happens to be all of $A_{4.}$ But in general, if $p = 2^{n}-1$ is a prime, the the semidirect product $EC$ is solvable, where $E$ is elementary Abelian of order $2^{n}$ and $C$ acts on $E$ as a Singer cycle in ${\rm GL}(n,2).$ Furthermore, in its action on its Sylow $p$-subgroups by conjugation, $EC$ is a permutation group of degree $p+1$ and contains $p$-cycles which don't commute. $\endgroup$ – Geoff Robinson Jun 18 '14 at 16:58
  • $\begingroup$ It won't be too surprising if Camille Jordan knew this result already (due to en.wikipedia.org/wiki/Jordan's_theorem_(symmetric_group)) $\endgroup$ – Dima Pasechnik Jun 19 '14 at 12:34
  • 1
    $\begingroup$ Maybe I am missing something, but I think a permutation group $G=\langle \sigma, \tau \rangle$ generated by two non-commuting $p$-cycles is primitive: If $\Delta$ is a block, then $\Delta \cap \operatorname{Mov} \sigma$ is empty or a block for $\langle \sigma \rangle$, and the same holds for $\tau$. By going through the different possibilities, I get that $\Delta$ must be trivial (or did I forget a case?). Of course, degrees p, p+1, p+2 remain. (I would be interested in a hint or sketch of your proof, by the way.) $\endgroup$ – Frieder Ladisch Jun 19 '14 at 15:20
  • 1
    $\begingroup$ @FriederLadisch: Need to prove G not p-solvable. Suppose oherwise. Can assume G = <x,y> for non-disjoint p-cycles& that deg G < 2p, G transitive. Hence <x> and <y> are Sylow. Op(G) = 1, Op'(G) is not 1.Then Hall-Higman type arguments get to G = <x>E, where E is elementary or special q-group some prime q ( NB x centralizes every proper x-invariant subgroup of E). H = point stabilizer, corefree. Can eliminate E non-Abelian with care as H meets Z(E) \leq Z(G) trivially in that case. Case E Abelian leads to H = <x>, [G:H] = p+1 = |E|, so p is Mersenne. $\endgroup$ – Geoff Robinson Jun 19 '14 at 16:04

Fixed several inaccuracies, many thanks to Frieder Ladisch for spotting them:

Let $x$ and $y$ be two non-commuting $p$-cycles, $G=\langle x,y\rangle$, and $G$ be considered as a transitive permutation group on the support $\Omega$ of $G$.

We show that either $\text{AGL}_1(\mathbb F_q)\le G$ for a Mersenne prime $q$, or $G$ is simple non-abelian.

Proof. $G$ is primitive (as pointed out by Frieder Ladisch already). This can be seen as follows: Let $\Delta$ be a block of a non-trivial block system. The action of the $p$-cycle $x$ on the block system is trivial, for otherwise $x$ would move $p\lvert\Delta\rvert>p$ points. On the other hand $G$ transitively moves the blocks, a contradiction.

Next we show that $G$ is doubly transitive. This follows from Burnside's classical theorem if $\lvert\Omega\rvert=p$. If $\lvert\Omega\rvert>p$, then the pointwise stabilizer in $G$ of the points fixed by $x$ is transitive (via $x$) on the remaining points. By Jordan's Theorem on primitive groups with Jordan sets we again see that $G$ is doubly transitive.

Let $N$ be a minimal normal subgroup of $G$. By Burnside, $N$ is either (a) elementary abelian and regular, or $N$ is (b) simple, primitive, and not regular. The Mersenne exceptions follow from looking at $p$-cycles in $\text{GL}_m(q)$ where $n=q^m$ for a prime $q$: Such a $p$-cycle fixes $q^r<q^m$ points, hence $p=q^r(q^{m-r}-1)$. We obtain $r=0$ and $q=2$. By Schur's Lemma, we can identify $\langle x\rangle$ with the multiplicative group of $\mathbb F_q$. This yields $\text{AGL}_1(\mathbb F_q)\le G$ as claimed. (If one uses Kantor's paper on Singer cycles, then one gets more precisely the possibilities $G=\text{AGL}_1(\mathbb F_q)$, $\text{A$\Gamma$L}_1(\mathbb F_q)$, and $\text{AGL}_m(\mathbb F_2)$. However, that paper relies on a wrong paper of Cameron/Kantor, see here.)

Now assume case (b). We show that $G=N$, so $G$ is actually simple. In order to do so, we show that $p$ divides the order of $N$. Note that $\lvert\Omega\rvert<2p$, so $p^2$ does not divide $\lvert N\rvert$, hence $x,y\in N$ in this case.

The case $p=\lvert\Omega\rvert$ is clear.

So $p<\lvert\Omega\rvert$ from now on. We let $\omega$ be a fixed point of $x$, and set $\Omega'=\Omega\setminus\{\omega\}$.

First suppose $p=\lvert\Omega\rvert-1$. The point stabilizer $N_\omega$ is a normal subgroup of $G_\omega$. As $G_\omega$ is transitive on $\Omega'$, all orbits of $N_\omega$ on $\Omega'$ have equal length. But $\lvert\Omega'\rvert=p$, so these orbit lengths are either $1$ or $p$. The former cannot hold, because then $N$ were regular on $\Omega$. Thus $N$ is doubly transitive on $\Omega$, so $p=\lvert\Omega\rvert-1$ divides $\lvert N\rvert$.

If $p<\lvert\Omega\rvert-2$, then $N=\text{Alt}(\Omega)$ by Jordan, so $G=N$ again.

If $p=\lvert\Omega\rvert-2$, then $G_\omega$ contains the cycle $x$ of length $p=\lvert\Omega'\rvert-1$ on $\Omega'$, so $G_\omega$ it is doubly transitive on $\Omega'$. The argument in the case $p=\lvert\Omega\rvert-1$ shows that $N_\omega$ is either doubly transitive on $\Omega'$, or regular. In the former case $p=\lvert\Omega'\rvert-1$ divides $\lvert N_\omega\rvert$, hence $G=N$ again. In the latter case, $N$ is sharply doubly transitive on $\Omega$, which implies that $N$ has a regular normal subgroup, contrary to $N$ being simple. (For this last step, simple counting suffices. One does not need Frobenius' theorem about the existence of Frobenius kernels.)

  • 1
    $\begingroup$ The case $|\Omega|=p$ is actually somewhat easier: if $N$ is a minimal subgroup of $G$, then $N$ is transitive and thus $x, y\in N$, so $G=N$ is simple. In the general "almost simple" case, is it clear that the order of the unique minimal normal subgroup $N$ is divisble by $p$ (and thus $G=N$ simple), without invoking Hall-Higman or similar arguments? Because this is what Geoff's argument shows: Either $G$ is simple, or $G \leq \operatorname{AGL}(m,2)$, where $p=2^m-1$. $\endgroup$ – Frieder Ladisch Jun 24 '14 at 14:26
  • 1
    $\begingroup$ Thanks, but I can't follow your argument in the case $n=p+1$. I think this is exactly the case where one needs something like Hall-Higman reduction to see that if $p$ does not divide $N$, then $N$ is not non-abelian simple. Also I don't know what you mean by "repeating the argument with $N$ instead of $G$" in case $p=n-2$. In that case, I have the following argument: if $N$ is $p'$, then $C_N(x)$ is transitive on the two elements fixed by $x$ (Lemma of Glauberman on coprime action), thus $N$ contains a 2-cycle, contradiction since $G\leq A_n$. $\endgroup$ – Frieder Ladisch Jun 24 '14 at 16:07
  • 1
    $\begingroup$ @GeoffRobinson: I rewrote the arguments. Hope the case $n=\lvert\Omega\rvert=p+1$ is clearer now. $\endgroup$ – Peter Mueller Jun 25 '14 at 8:37
  • 1
    $\begingroup$ @Peter Müller: I find your argument for $p= |\Omega|-1$ very simple and elegant. In case $p = |\Omega|-2$ I think one still needs to say why $N_{\omega}$ can not be regular on $\Omega'$, for example, because otherwise $N$ would be a doubly transitive Frobenius group, thus not simple. $\endgroup$ – Frieder Ladisch Jun 25 '14 at 11:18
  • 1
    $\begingroup$ (And in the Mersenne case, also $G= \operatorname{ASL}(m,2)$ and $G= \operatorname{A\Gamma L}(1,2^m)$ is possible.) $\endgroup$ – Frieder Ladisch Jun 25 '14 at 11:21

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.