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Next: Min size of distinguishing Up: Discrete Math, 13th Problem Previous: Decision tree complexity

Large primitive groups

For large $ n$, the largest four primitive permutation groups are $ S_n$ and $ A_n$, of order about $ n!$, and $ S_k^{(2)}$ (for $ n=\binom{k}{2}$) and $ S_k\wr S_2$ (for $ n=k^2$), of order about $ e^{c\sqrt n\log n}$. The classification of finite simple groups allows us to show that these are the largest and even to list the largest down to size about $ e^{\log^2 n}$. We can do reasonably well with elementary means.

Theorem 2.1   Assume $ G<S_n$, $ A_n \nless G$, and $ G$ is primitive.
  1. (Bochert c.1890) $ \vert G\vert\le \frac{n!}{(n+1)/2)!}\approx e^{\frac n2\log n}$.
  2. (Wielandt, Praeger-Saxl, 1980) $ \vert G\vert<4^n$.
  3. (Babai, 1981) If $ G$ is not doubly transitive, $ \vert G\vert<e^{4\sqrt n\log^2n}$ (using almost only graph theory).


\begin{exercise}Doubly transitive implies primitive.
\end{exercise}

The following theorem si proved using the classification of finite simple groups; one can get close by elementary means.

Theorem 2.2   If $ G\le S_n$, $ G\not\ge A_n$ is doubly transitive then $ \vert G\vert<n^{1+\log_2 n}.$


\begin{exercise}
% latex2html id marker 1103
Verify this bound for $\PSL$\ and $\AGL$, acting on
projective and affine spaces, resp.
\end{exercise}

Remarks about symmetry and regularity: symmerty conditions are given in terms of automorphisms; regularity conditions in terms of numerical parameters. Symmetry condition imply regularity conditions (e.g., vertex-transitivity is a symmetry condition, which implies that the graph is regular, a regularity condition). The converse is seldom true. We shall define regularity conditions on a family of edge-colored digraphs which capture some combinatorial consequences of primitive group action. Using this translation, we shall prove a combinatorial result which implies a nearly optimal upper bound on the order of uniprimitive (primitive but not doubly transitive) permutation groups.

Picture of $ D_6$. % latex2html id marker 2557
$ R_0 = \Delta = \{(x,x) \mid x \in \Omega\}$, diagonal. % latex2html id marker 2559
$ \Omega \times \Omega = R_0 \cup R_1 \cup \dots \cup R_{r-1}$. $ r = $# colors = # orbits of $ G$ on % latex2html id marker 2565
$ \Omega \times \Omega$. $ D_6$ has rank $ 4$, $ r = 4$. In this case, all orbitals are self-paired.

Definition 2.3   An orbital $ \Gamma$ of a permutation group % latex2html id marker 2576
$ G\le \Sym(\Omega)$ is an orbit of $ G$ on the set of ordered pairs ( % latex2html id marker 2580
$ \Gamma\subset\Omega\times\Omega$). $ \Gamma$ is self-paired when $ \Gamma=\Gamma^{-1}$ (i.e., for $ (x,y)\in \Gamma$ there exists $ \sigma\in G$ such that $ x^{\sigma}=y$ and $ y^{\sigma}=x$). The rank $ r$ of a permutation group is the number of its orbitals.


\begin{exercise}
% latex2html id marker 1105If $G$\ is doubly transitive, then...
...\rm rk}\nolimits (G) = 2$.
What do the two classes correspond to?
\end{exercise}

Definition 2.4   COHERENT CONFIGURATION of rank $ r$:
% latex2html id marker 2599
$ \ensuremath{\mathfrak{X}}= (\Omega; R_0, \dots, R_{r-1})$, % latex2html id marker 2601
$ R_i \subseteq \Omega \times \Omega$.
% latex2html id marker 2603
$ \Omega \times \Omega = R_0 {\mathbin{\dot{\cup}}}\dots {\mathbin{\dot{\cup}}}R_{r-1}$.
% latex2html id marker 2605
$ X_i = (\Omega, R_i)$, $ i$'th color digraph, called a constituent digraph. The color of a pair $ x,y$ is defined as $ c(x,y) = i$ if $ (x,y) \in R_i$.
To be coherent, the following 3 axioms must be satisfied:
A1:
The diagonal is $ \Delta = R_0 {\mathbin{\dot{\cup}}}\dots {\mathbin{\dot{\cup}}}R_{i_0 - 1}$. Equivalently, $ c(x,x) = c(y,z) \Rightarrow y=z$.
A2:
% latex2html id marker 2622
$ (\forall i)(\exists j)(R_j = R_{i}^{-1})$. Terminology: $ R_i$ is self-paired if $ R_i = R_{i}^{-1}$, i.e., $ X_i$ is undirected.
A3:
% latex2html id marker 2630
$ (\exists p_{i,j,k})(\forall (x,y) \in R_i)
(\char93  \{z \mid c(x,z) = j, c(z,y) = k\} = p_{i,j,k})$

Definition 2.5   For % latex2html id marker 2636
$ G \le \Sym(\Omega)$, % latex2html id marker 2638
$ \ensuremath{\mathfrak{X}}(G) := (\Omega;$   orbitals$ )$. We refer to these as ``the group case.''


\begin{exercise}
% latex2html id marker 1124$\ensuremath{\mathfrak{X}}(G)$\ is a coherent configuration.
\end{exercise}


\begin{exercise}
% latex2html id marker 1128$G \le \mathop{\rm Aut}(\ensuremat...
...{\mathfrak{X}})$\ if $(\forall x,y)(c(x,y) = c(x^{\pi}, y^{\pi}))$\end{exercise}

Remark 2.6   There exist coherent configurations without a group. In fact, there are exponentially many rank-3 coherent configurations with no automorphisms.

Well, we always lose in translation. The question is how much.


\begin{exercise}
% latex2html id marker 1136The number of $x \to \dots \to y$\...
... $y$\ of length 4 are
colored red, blue, purple, blue (in order)?
\end{exercise}

Definition 2.7   $ \ensuremath{\mathfrak{X}}$ is homogeneous if $ R_0 = \Delta$ (i.e., $ (\forall x,y)(c(x,x) = c(y,y))$).


\begin{exercise}
% latex2html id marker 1140$\ensuremath{\mathfrak{X}}(G)$\ is homogeneous $\Longleftrightarrow $\ $G$\ is transitive.
\end{exercise}


\begin{exercise}
% latex2html id marker 1144If $\ensuremath{\mathfrak{X}}$\ is...
..., then every weak component of
each $X_i$\ is strongly connected.
\end{exercise}


\begin{exercise}
% latex2html id marker 1148If $\ensuremath{\mathfrak{X}}$\ is...
...not depend on $x$).
So $X_i$\ is Eulerian, and indeed is regular.
\end{exercise}

By the way, $ \sum_{i=0}^{r-1} \rho_i = n$, since every vertex is connected to every other (including itself) in the graph $ \cup X_i$, whose edge set contains all $ n^2$ ordered pairs.

Definition 2.8   $ \ensuremath{\mathfrak{X}}$ is a primitive coherent configuration if $ \ensuremath{\mathfrak{X}}$ is homogeneous and ALL constituent digraphs $ X_i$, $ i \ge 1$ are connected.


\begin{exercise}
% latex2html id marker 1160$\ensuremath{\mathfrak{X}}(G)$\ is primitive $\Longleftrightarrow $\ $G$\ is primitive. (DO!!!)
\end{exercise}

Definition 2.9   $ \ensuremath{\mathfrak{X}}$ is uniprimitive coherent configuration if $ \ensuremath{\mathfrak{X}}$ is primitive and rank $ \ge 3$.


\begin{exercise}
% latex2html id marker 1168$\ensuremath{\mathfrak{X}}$\ is un...
...ow $\ $G$\ is uniprimitive (primitive but not
doubly transitive).
\end{exercise}

% latex2html id marker 2671
$ G \le \Sym(\Omega)$, % latex2html id marker 2673
$ \Psi \subseteq \Omega$. Look at the pointwise stabilizer, $ G_{\Psi}$, $ = \cap_{x \in \Psi} G_x$. If $ G_{\Psi} = \{1\}$, then $ \vert G\vert \le n^{\vert\Psi\vert}$, in fact $ \vert G\vert \le n(n-1) \dots (n-\vert\Psi\vert+1)$. Call such a $ \Psi$ a ``fixing set.''

We shall prove, using only elementary graph theoretic arguments, that

Theorem 2.10   If $ G$ is uniprimitive, then $ \vert G\vert < e^{4 \sqrt{n} (\ln n)^2}$.

Lemma 2.11   If $ G$ is uniprimitive, then % latex2html id marker 2695
$ (\exists \Psi \subseteq \Omega)
(\vert\Psi\vert \le 4 \sqrt{n} \ln n$    and $ G_\Psi = \{1\})$.

Examples: How large is the smallest fixing set for various classes of permutation groups?

Definition 2.12   $ z$ distinguishes $ x$ and $ y$ if $ c(x,z) \ne c(y,z)$. $ D(x,y) = \{z \mid c(x,z) \ne c(y,z) \}$ is the distinguishing set for $ x,y$.


\begin{exercise}
% latex2html id marker 1172If $\ensuremath{\mathfrak{X}}= \en...
...rbit of $G_z$. (Obvious, because the group preserves the colors.)
\end{exercise}

Definition 2.13   A distinguishing set of $ \ensuremath{\mathfrak{X}}$ is any set % latex2html id marker 2714
$ \Psi \subseteq \Omega$ such that $ (\forall x \ne y)(\Psi \cap D(x,y) \ne \emptyset)$. In other words, for every pair $ x,y$, $ \Psi$ contains an element which distinguishes them.


\begin{exercise}
% latex2html id marker 1180For $\ensuremath{\mathfrak{X}}= \e...
...$\ is a distinguishing set,
then $\Psi$\ is a fixing set for $G$.
\end{exercise}

Theorem [*] will follow from the following result.

Theorem 2.14   If $ \ensuremath{\mathfrak{X}}$ is a uniprimitive coherent configuration, then there exists a distinguishing set $ \Psi$ such that $ \vert\Psi\vert < 4 \sqrt{n} \ln n$.

This will be an immediate consequence of the following. From now on, let us always assume $ \ensuremath{\mathfrak{X}}$ is a uniprimitive coherent configuration.

Theorem 2.15 (Main technical theorem)   For every $ x,y$, $ \vert D(x,y)\vert \ge \sqrt{n}/2$.

Proof. [Main technical theorem $ \Rightarrow $ Theorem [*]]. Pick $ u_1, \dots, u_m$ at random, and hope that we picked enough to hit each $ D(x,y)$.


$\displaystyle \Prob(D(x,y)$    not hit$\displaystyle )$ $\displaystyle =$ $\displaystyle \left(1 - \frac{\vert D(x,y)\vert}{n} \right)^m$  
  $\displaystyle \le$ % latex2html id marker 2755
$\displaystyle \exp\left (-\frac{\vert D(x,y)\vert m}{n} \right).$  

Hence, by the Union Bound,
% latex2html id marker 2758
$\displaystyle \Prob((\exists x,y)(D(x,y)$    not hit$\displaystyle ))$ $\displaystyle <$ % latex2html id marker 2763
$\displaystyle \binom{n}{2}
\exp\left(-\frac{{D_{\mathrm{min}}}m}{n} \right)$  
  $\displaystyle <$ % latex2html id marker 2767
$\displaystyle \exp\left(-\frac{{D_{\mathrm{min}}}m}{n} + 2 \ln n\right),$  

where $ {D_{\mathrm{min}}}= \min_{x \ne y} \vert D(x,y)\vert$.

For this, it is sufficient to show

% latex2html id marker 2771
$\displaystyle \exp\left(\frac{{D_{\mathrm{min}}}m}{n} + 2 \ln n\right) \le 1
$

or equivalently

$\displaystyle \frac{{D_{\mathrm{min}}}m}{n} + 2 \ln n \le 0
$

which follows from

$\displaystyle m \ge \frac{2 n \ln n}{{D_{\mathrm{min}}}} \le 4 \sqrt{n} \ln (n) =: m.
$

The last inequality used the Main technical theorem, which lower bounds $ {D_{\mathrm{min}}}$. $ \qedsymbol$


next up previous
Next: Min size of distinguishing Up: Discrete Math, 13th Problem Previous: Decision tree complexity
Varsha Dani 2003-08-15