# Permutation Groups Containing non-commuting $p$-cycles

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).

• So you say e.g. that ${\rm A}_4$ is solvable because $3$ is a Mersenne prime? – Stefan Kohl Jun 18 '14 at 15:36
• 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. – Geoff Robinson Jun 18 '14 at 16:58
• It won't be too surprising if Camille Jordan knew this result already (due to en.wikipedia.org/wiki/Jordan's_theorem_(symmetric_group)) – Dima Pasechnik Jun 19 '14 at 12:34
• 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.) – Frieder Ladisch Jun 19 '14 at 15:20
• @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. – 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.)

• 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 $p=2^m-1$. – Frieder Ladisch Jun 24 '14 at 14:26
• 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 $G\leq A_n$. – Frieder Ladisch Jun 24 '14 at 16:07
• @GeoffRobinson: I rewrote the arguments. Hope the case $n=\lvert\Omega\rvert=p+1$ is clearer now. – Peter Mueller Jun 25 '14 at 8:37
• @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. – Frieder Ladisch Jun 25 '14 at 11:18
• (And in the Mersenne case, also $G= \operatorname{ASL}(m,2)$ and $G= \operatorname{A\Gamma L}(1,2^m)$ is possible.) – Frieder Ladisch Jun 25 '14 at 11:21