Iwasawa's lemma

Lemma (Iwasawa) If $(G,\Omega)$ is a primitive permutation group with $G$ perfect and for some $\alpha \in \Omega$, $G_\alpha$ has a normal abelian subgroup $A$ whose conjugates generate $G$, then $G$ is simple.
proof: Suppose $1 \not = N \unlhd G$, we will gradually show that $N$ must be the whole group. By primitivity $N$ is transitive on $\Omega$ and $G_\alpha$ is a maximal subgroup, so $N \not \le G_\alpha$ and $N G_\alpha = G$. Any $g$ may be written $n x$ for some $n \in N$, $x \in G_\alpha$ so $A^g = A^{nx} = A^x \le AN$ and these conjugates cover the whole group so $AN = G$. Now $G/N \simeq A/(A \cap N)$ is abelian, but $G$ is perfect so $N = G$.

Projective Spaces and Groups

We aim to "fix" the following two issues: GL and SL are not 2-transitive because they can't linearly dependent vectors to linearly independent ones. SL is not simple (even though it's perfect) because it has a center. Let $V=V_n(q)$ and $n \ge 2$ throughout.

We define an equivalence relation $R$ on $V^\#$ by $vRw$ iff $v = \lambda w$ for some nonzero $\lambda \in \mathbb F_q$.

Definition We then have the projective space $\mathbb P(V)$ of projective vectors. Write $\mathbb P^{n-1}(q)$.

For a subspace $U \subseteq V$ the set of equivalence classes (projective vectors) $[U]$ (the image of $U^\#$) is a subspace of $\mathbb P(V)$ so it inherits geometric structure. The dimension of $[U]$ is the dimension of $U$ minus 1. A point is a class $[v]$ for some vector $v$, and a line is the projective class of a 2D subspace.

Given $g \in GL(V)$, $v \in V^\#$ and $\lambda \in \mathbb F_q^\#$ we have $(\lambda v)g = \lambda (vg) \in [vg]$ so we can (well) define an action by $[v]g = [vg]$. In this way $GL(V)$ and $SL(V)$ act on $\mathbb P(V)$, but not faithfully.

Lemma Let $G$ be $GL(V)$ or $SL(V)$, the kernel of the action of $G$ on $\mathbb P(V)$ is $Z(G)$.
proof: From the previous post we have seen what $Z(G)$ is: scalar multiples of the identity. If $gs = 1$ then clearly $g$ acts trivially on $\mathbb P(V)$. Conversely if $g \in GL(V)$ is in the kernel of the action then $[vg]=[v]$ for all $[v] \in \mathbb P(V)$, then for every vector $V$ we have $v g = \lambda_v v$ for some $\lambda_v \in \mathbb F^\#$ and the proof concludes in the same way as before.

Definition The projective general linear group and projective special linear group are defined by $PGL(V) = GL(V)/Z(GL(V))$ and $PSL(V) = SL(V)/Z(SL(V))$. They have faithful action on $\mathbb P(V)$. (Note: PSL might not be a subgroup of PGL anymore).

We call $[g] \in PGL$ or $PSL$ the image (???) if $g \in GL$ or $SL$ if $[v][g] = [vg]$ for $v\in \mathbb P(V)$. $PGL_n(q)$ and $PSL_n(q)$ act on $\mathbb P^{n-1}(q)$ by the order calculations in the previous post we find $|PGL_n(q)| = q^{n(n-1)/2}\prod_{i=2}^n(q^i-1)$ and $|PSL_n(q)| = \frac{q^{n(n-1)/2}}{(n,q-1)}\prod_{i=2}^n(q^i-1)$. Thus $|PGL_2(q)| = (q+1)q(q-1)$.

Proposition The permutation group $PGL_2(q)$ is sharply 3-transitive on $\mathbb P^1(q)$.
proof: For $g = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \in GL_2(q)$ and $v \in V_2(q)^\#$, $vg = (a \lambda_1 + c \lambda_2, b \lambda_1 + d \lambda_2)$. These work out exactly as the mobius transformations when regarding $\mathbb P^1(q)$ as $\mathbb F_q \cup \{\infty\}$. We have already shown this one point extension is genrated by the mobius transforms!

Proposition Both $PGL(V)$ and $PSL(V)$ act 2-transitively on $\mathbb P(V)$.
proof: Given $([e_1],[e_2])$, $([e'_1],[e'_2])$ distinct then extend to a basis and find a map between them (why doesn't this work for all n-transitivity?)

Transvections

Assume $n \ge 2$ throughout,

Definition A linear functional on $V$ is a linear map from $V$ to $\mathbb F_q$. The set of these is the dual space $V^*$. Given $f \in V^*$ write $V_f$ for the kernel of $f$, if $f \not = 0$ then $V_f$ is a subspace of dimension $n-1$ (any such space of codimension 1 we call a hyperplane).

Lemma If you have $f,f' \in V$ with the same hyperplane then there exists $\lambda$ such that $f' = \lambda f$.
proof: The result is clear if $V_f=V$ assume not so $f,f'\not = 0$. Take $v \in V \setminus V_f$ (i.e. not in the common hyperplane) so $vf, vf' \not = 0$ and put $\lambda = (vf')(vf)^{-1}$ then $f' - \lambda f \in V^*$ and its kernel is contained in $\langle V_f, v \rangle = V$ so it equals zero.

Definition A linear automorphism $\tau \in GL(V)$ is called a transvection with direction $d \in V^\#$ if $\tau$ fixes $d$ and $$v \tau - v$$ is a scalar multiple of $v$ for all $v$.

Clearly $1$ is a transvection of any direction.

Lemma If $\tau$ is a transvection with direction $d$ then the vectorspace $\operatorname{fix}(\tau)$ is a hyperplane containing $d$.
proof: Define $f : V \to \mathbb F_q$ by $(vf)d = v \tau - v$ i.e. $f$ is that scalar multiple which a transvection defines. Since $\tau$ is linear $f \in V^*$. The result is clear for $\tau = 1$ and if not $f \not = 0$ so $d \in \operatorname{fix}(\tau) = V_f$.

Corollary Any transvection can be written as $\tau_{f,d}$ mapping $v$ to $v + (vf)d$ for some $f \in V^*$, $d \in (V_f)^\#$.

Lemma For $f,f' \in V^*$, $d \in V^\#$, $g \in GL(V)$ we have

• $\tau_{f,g}^g = \tau_{g^{-1}\circ f,dg}$
• $\tau_{f,d} \tau_{f',d} = \tau_{f+f',d}$

proof: first we need to check $(dg)(g^{-1} \circ f) = 0$ and $d(f+f')=0$ for the notation to be meaningful. Now the results follow by computation.

Definition For a direction $d \in V^\#$ set $\mathscr T(d) = \{\tau_{f,d}|f \in V^*, d \in V_f\}$ and $\mathscr T$ the union over all directions (all transvections).

Proposition $\mathscr T^\#$ is a single conjugacy class in $GL(V)$ and lies in $SL(V)$. If $n\ge 3$ then $\mathscr T^\#$ is a single conjugacy class in $SL(V)$.
proof: By the calculation lemma previous, we know that $\mathscr T^\#$ is closed under conjugation. Let $\tau_{f,d},\tau_{f',d'} \in \mathscr T^\#$ and write $e_1=d,e'_1=d'$ then take bases $e_1,\ldots,e_{n-1}$ and $e'_1,\ldots,e'_{n-1}$ of the hyperplanes $V_f,V_{f'}$, choose $e_n,e'_n$ such that $e_n f = e_n' f' = 1$. Now we have bases for $V$ so there is a GL map $g$ from one to the other.
For $i < n$ we have $e'_i(g^{-1} \circ f) = e_i f = 0$ so $V_{g^{-1} \circ f}$ is the space spanned by $\langle e'_1,\ldots,e'_{n-1} \rangle = V_{f'}$ so they are scalar multiples of each other, let $\lambda$ be such that $f' = \lambda (g^{-1} \circ f)$ and $$1 = e'_n f' = \lambda e'_n (g^{-1} \circ f) = \lambda e_n f = \lambda$$ so since $dg = d'$ it follows that $\tau_{f,d}^g = \tau_{f',d}$!
If they are all conjugate they all have the same determinant $\delta$, now $\det(\tau_{f,d}\tau_{f',d}) = \det(\tau_{f+f',d})$ so $\delta^2 = \delta$ proves they lie in $SL$.
For $n \ge 3$ we can use the $\mu$ trick as before to get them in $SL$.

Proposition If $d \in V^\#$ then $\mathscr T(d)$ is an abelian normal subgroup of the stabilizer $SL(V)_d$; $\mathscr T(d)$ are all conjugates in $SL(V)$.
proof: Certainly elements of $\mathscr T(d)$ stabilize $d$, by the computational lemma before we see that it is an abelian group (commutative and closed, therefore has identity and inverses). If $g \in SL(V)$ with $dg=d$ then $\mathscr T(d)^g = \mathscr T(d)$ (by the previous) so $\mathscr T(d) \unlhd SL(V)_d$. Given $d,d' \in V^\#$ by the transitivity lemma (that requires dimension > 1) exists $g \in SL(V)$ which takes $d$ to $d'$ then $\mathscr T(d)^g = \mathscr T(d')$.

Proposition The set $\mathscr T$ generates $SL(V)$.
proof: Elementary matrices.

Definition A group if perfect if $G' = G$. This is equivalent to there being no nontrivial abelian quotients: Clearly if $G/[G,G] \not = 1$ then $G\not = [G,G]$. Conversely if $G/N \simeq A = \{Ng\}$ then we always have $Ngg' = Ng'g$ i.e. there is some $n$ such that $gg' = ng'g$. So every commutator $[g,g']$ is an element of $N$.

Proposition If $n \ge 2$ the group $SL_n(q)$ is perfect provided $(n,q)$ isn't $(2,2)$ or $(2,3)$.
proof: We just need to show that each $\tau \in \mathscr T^\#$ is a commutator since that set generates our group. If $\tau$ has direction $d$ take some $\sigma \in \mathscr T(d)^\#$ not equal to $\tau^{-1}$, then $\sigma \tau \in \mathscr T(d)^\#$ so there is a $g \in SL_n(q)$ that conjugates $\sigma \tau = \sigma^g$, whence $\tau = \sigma^{-1} g^{-1} \sigma g= [\sigma,g]$.
For $n=2$ we will use $2 \times 2$ matrices directly, in some basis $\tau = \begin{pmatrix} 1 & \gamma \\ 0 & 1 \end{pmatrix}$ (nonzero $\gamma$), now for any nonzero $\lambda$ and $\mu \in \mathbb F_q$ we have $$\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}\begin{pmatrix} 1 & \mu \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \lambda^{-1} & 0 \\ 0 & \lambda \end{pmatrix}\begin{pmatrix} 1 & -\mu \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & \mu(\lambda^2-1) \\ 0 & 1 \end{pmatrix}$$ so for $q>3$ take $\lambda \not = 0,1,-1$ then $\lambda^2-1\not = 0$ so let $\mu = \gamma(\lambda^2-1)^{-1}$.

Proposition

• $Z(GL(V)) = \{\lambda 1 | \lambda \in \mathbb F_q^\# \}$
• $Z(SL(V)) = \{\lambda 1 | \lambda \in \mathbb F_q^\#, \lambda^n = 1\}$

proof: Certainly these are contained in the center, suppose $g \in Z(GL(V))$ or $Z(SL(V))$ then for each nonzero vector $d \in V^\#$ take a transvection $\tau_{f,d} \in \mathscr T(d)$, this should be unaffected by conjugation but we know $\tau_{f,d}^g = \tau_{g^{-1} \circ f,dg}$ so $dg = \lambda_d d$. Then $$\lambda_d d + \lambda_v d = (d+v)g = \lambda_{d+v}(d+v) = \lambda_{d+v}d + \lambda_{d+v}v$$ so $\lambda_d = \lambda_v = \lambda$ and $g = \lambda \cdot I$. For $SL$ we also require $\det(g) = \lambda^n = 1$. This tells us that $Z(GL_n(q)) \simeq \mathbb F_q^\#$ and $Z(SL_n(q)) \simeq C_{(n,q-1)}$.

Finite Fields and Finite Vector Spaces

Definition An Affine transformation of $\mathbb F_q$ is a map $f_{a,b}$ taking $\lambda$ to $a \lambda + b$ where $a \in \mathbb F_q^\#$ is nonzero. The group of such maps is called $A(\mathbb F_q)$.

Proposition $A(\mathbb F_q)$ is sharply 2-transitive of order $q(q-1)$.
proof: Let $\alpha,\beta$ distinct, the system of equations $\alpha = a 0 + b$, $\beta = a 1 + b$ has a unique solution.

Corollary By a general lemma about sharply 2-transitive groups this group must have a regular characteristic subgroup, this group is $\{f_1,b|b\in\mathbb F_q\}$.

Proposition $A(\mathbb F_q)$ has a one point extension which is sharply $3$-transitive of degree $q+1$.
proof: This is a straightforward application of the one point extension theorem, adjoin $\infty$ to $\mathbb F_q$ and define $x$ on $\mathbb F_q \cup \{\infty\}$ to swap $0$ and $\infty$ and invert all other elements $\lambda x = \lambda^{-1}$. Clearly $x^2=1$. Let $G_0 = \{f_{a,0}\mid a \in \mathbb F_q^\# \}$ and note $f_{a,0}^x$ fixes $0$ and $\infty$ while for $\lambda \in \mathbb F_q^\#$ we $f_{a,0}^x = f_{a^{-1},0}$ so $G_0^x = G_0$. Finally a system for the double cosets is given by just $1$ and any other element e.g. $f = f_{-1,1}$ will do (so $\lambda f = 1 - \lambda$ and $f^2=1$). We see that $xf$ acts on the $\infty, 1, 0$ by cycling them and for the remaining elements $\lambda (xf)^3 = 1 - \frac{1}{1-\frac{1}{\lambda}}=\lambda$ and (apparently...) $x^f = f^x \in GxG$ so we have a one point extension.

Definition $V_n(q)$ is the $n$-dimensional vector space over $\mathbb F_q$, clearly $|V_n(q)|=q^n$.

Definition If $V$  is a vector space then a linear automorphism of $V$ is a bijective linear map $V \to V$. The group of these is called the general linear group $GL(V)$ or $GL_n(q)$ when $V=V_n(q)$.

Definition The special linear group $SL(V)$ of linear automorphisms of determinant 1.

Lemma If $V$ is a vector space over $\mathbb F_q$ then $SL(V)$ is a normal subgroup of $GL(V)$ and the index is $q-1$: $|GL(V):SL(V)|=q-1$.
proof: SL is just the kernel of the surjective determinant map from GL to $\mathbb F_q^\times$. As a consequence $GL/SL \simeq \mathbb F_q^\#$ so $|GL:SL| = q-1$.

Lemma The group $GL(V)$ acts transitively on $V^\#$ and if the dimension of $V$ is $> 1$ the same is true of $SL(V)$.
proof: Take two nonzero vectors $e_1,f_1$ then to get a map between them choose bases $e_1,\ldots,e_n$ and $f_1,\ldots,f_n$ this gives $g \in GL(V)$ mapping between them. If $n > 1$, since we don't necessarily have $\det(g)=1$ let $\det(g)=\mu$ and replace $e_n$ by $\mu^{-1} e_n$. Now $g'$ mapping between these bases has determinant $1$. (did I get this right?)

Proposition $$|GL_n(q)| = q^{n(n-1)/2} \prod_{i=1}^n (q^i-1)$$ and $$|SL_n(q)| = q^{n(n-1)/2} \prod_{i=2}^n (q^i-1).$$
proof: $GL_n(q)$ acts regularly on ordered bases of $V_n(q)$, so the size of $GL_n(q)$ is equal to the number of ordered bases: $q^n-1$ choices for the first element, and having chosen $e_1,\ldots,e_i$ (which spans a $q^i$ sized space) already there are $q^n-q^i$ choices for the next. The size of $SL$ comes from the lemma before the previous.

Sharply t-transitive groups

Sharply transitive groups are the smallest possible transitive groups, given $t \in \mathbb N$ a $t$-transitive group $G$ is called sharply t-transitive if $G_{\alpha_1 \alpha_2 \ldots \alpha_t} = 1$ for distinct $\alpha$s. Equivalently, there is exactly one group element that takes $(\alpha_1,\cdots,\alpha_t)$ to any other triple of distinct symbols.

Theorem If $G$ is a $t$-transitive group of degree $n$ (i.e. it acts on $n$ symbols) then it is sharply $t$-transitive iff $|G| = n (n-1)\cdots(n-t-1)$.
proof: For $t=1$ this is the regular action. For $t+1$ any $G_\alpha$ will be $t$-transitive and $|G|=n|G_\alpha|$ since the orbit of $\alpha$ is the whole of $\Omega$, conversely every stabilizer $G_\alpha$ will have the same order - and there are $n$ of them so each $|G_\alpha| = |G|/n$ is $t-1$-transitive so $G$ is $t$-transitive.

Example $S_n$ is sharply $n$-transitive in the natural actions, it is also $n-1$-sharply transitive. Note this isn't a contradiction from the group order result because $|G|=|G|\cdot 1$. Intuitively what's happening is if we choose exactly where $n-1$ symbols go, then it's already decided where the last one must go.

Proposition If $G$ is sharply $2$-transitive of degree $n$ then $G$ has a regular characteristic subgroup which is an elementary abelian $p$-group for some prime $p$ (so $n$ is power of $p$).
proof: Given $G$, let $K$ be the union of the fixed point free elements and 1. We will show it is a group. Since for any distinct $\alpha,\beta \in \Omega$, $G_\alpha \cap G_\beta = G_{\alpha \beta} = 1$ so $G$ is the disjoint union $K \sqcup \bigsqcup_{\alpha \in \Omega} G_\alpha^\#$, $|G_\alpha^\#| = \frac{|G|}{n}-1$ (using $n = |G|/|G_\alpha|$ from the orbit-stab. theorem) so $|K|=n$. Take $\alpha,\beta \in \Omega$ distinct and choose $k \in K^\#$ such that $\alpha k \not = \alpha$. As $G$ is $2$-transitive there exists $g \in G_\alpha$ with $(\alpha h) g = \beta$ then $h^g$ has the same order and fixedpoints as $h$ i.e. none, therefore its in $K$, and $\alpha k^g = \beta$ so $K$ is a transitive "set". For any $\alpha \in \Omega$ the map $K \to \Omega$ given by $k \mapsto \alpha k$ is surjective and hence bijective. Now take take distinct $x,y \in K$ we know $\alpha x \not = \alpha y$ so $\alpha x y^{-1} \not = \alpha$ for all $\alpha$, so $x y^{-1} \in K$. This proves $K$ a subgroup!
Given $k \in K^\#$ all its cycles on $\Omega$ must all have the same length since otherwise some non-identity power of $k$ would have fixed points. So the order of $k$ divides $n$. Similarly for elements of $G_\alpha^\#$ the same argument tells us all its cycles on $\Omega \setminus \{\alpha\}$ have the same length, so its order divides $n-1$. Hence $K$ consists of the elements of $G$ of order dividing $n$ - a property preserved by conjugation - therefore it's a characteristic subgroup. By the corollary from the section on regular normal subgroups it is an elementary abelian $p$-group.

One-point extensions

Reversing the idea of a point stabilizer Let $(G,\Omega)$ be a transitive permutation group, take a point $\omega \not\in \Omega$ and form $\Omega^+ = \Omega \cup \{\omega\}$, extend the action by $\omega g = \omega$ for all $g \in G$:

Definition A one-point extensionof $(G,\Omega)$ is a transitive permutation group $(G^+,\Omega^+)$ with $(G^+)_\omega = G$

By the stabilizer-orbit theorem $|G_+| = |G|(|\Omega|+1)$. If $G^+$ is $t$-transitive then G is $(t-1)$-transitive.

Example $S_n$ and $A_n$ have one point extensions $S_{n+1}$ and $A_{n+1}$.
Non-example $D_8$ doesn't have one by Sylow theory.

Take $\alpha \in \Omega$, we know the rank $r$ of $G$ equal to the number of double cosets in $G$. For $g_1,\ldots,g_r \in G$ we have a complete representation of the double coset system iff $\alpha g_1,\ldots,\alpha g_r$ is a complete set of representatives of $G_\alpha$-orbits. wlog take $g_1 = 1$, if $(G^+,\Omega^+)$ is a one point extension of $(G,\Omega)$ then it is 2-transitive so it has a primitive action and hence the point stabilizer $G$ is a maximal subgroup of $G^+$: for any $x \in G^+ \setminus G$ we must have $\langle x, G \rangle = G^+$. We can wlog choose $x$ to interchange $\alpha$ and $\omega$.

Theorem Let $(G,\Omega)$ be a transitive permutation group of rank $r$ for $\alpha \in \Omega$, let $g_1=1,g_2\ldots,g_r$ be a complete set of representatives of the double coset system. Take $\omega \not\in \Omega$ and form $\Omega^+ = \Omega \cup \{\omega\}$. Take $x \in S_{(\Omega^+)}$ (so some permutation from the symmetric group) with $\alpha x = \omega$, $\omega x = \alpha$ and set $G^+ = \langle x, G \rangle$ then $(G^+,\Omega^+)$ is a one point extension iff:

1. $x^2 \in G_\alpha$
2. $(G_\alpha)^x = G_\alpha$
3. $g_i^x \in GxG$ for all $i > 1$.

proof: ($\Rightarrow$) (1) definition of $x$ (2) easy (3) As $x \not\in G$ the 2-transitivity of $G^+$ means that $G^+ = G \cup GxG$ (any permutation that can't be done with $g$ can be done with $x$ in the middle), for $2 \le i \le r$ we have $g_i \in G\setminus G_\alpha$ so $\omega g_i^x = \alpha g_i x \not = \alpha x = \omega$ thus $g_i^x \not \in (G^+)_\omega = G$ so $g_i^x \in GxG$. ($\Leftarrow$) Set $H=G \cup GxG$as a set, we will show its a group $H \le S_{\Omega^+}$, to show it's a group we need $HH = H = H^{-1}$. Since $G^{-1}= G$ and $(gxg')^{-1} = g'^{-1} x (x^2)^{-1} g^{-1} \in GxG$ . By (2) $x G_\alpha x= G_\alpha$ so for $i \ge 2$ we have $$x G_\alpha g_i G_\alpha x = G_\alpha (x g_i x) G_\alpha = G_\alpha x^2 g_i^x G_\alpha \subseteq GxG$$ so $$xGx = xG_\alpha x \cup \bigcup_{i=2}^r x G_\alpha g_i G_\alpha x \subseteq G_\alpha \cup GxG \subseteq H.$$ Now $HH = G \cup GxG \cup GxGxG \subseteq H \cup GHG = H$, we have shown it's a group! Now $G^+ = \langle G,x\rangle = H$ and for all $g,g'$ take $\omega (g x g') = \omega x g' = \alpha g' \not = \omega$ so $(G^+)\omega = G$.

Multiple-transitivity

Consider actions of rank 2 (meaning that there are two suborbits of $G_\alpha$), $\Omega \setminus \{\alpha\}$ forms a single $G_\alpha$ orbit (because one its other orbit is $\alpha G_\alpha = \{\alpha\}$). $G$ acts on $\Omega^2$ non-transitively because $g(\beta,\beta) = (\gamma,\gamma)$ but if we define the diagonal $Delta = \{(\beta,\beta)\in \Omega\}$ this is a single orbit due to transitivity and so $\Omega^2 \setminus \Delta$ is the interesting part:

Lemma The action of $G$ on $\Omega$ is of rank 2 iff $\Omega^2 \setminus \Delta$ is a single orbit.
proof: If the action has rank 2 then let $(\beta_1,\beta_2), (\gamma_1,\gamma_2)$ lie off the diagonal and pick $x,y \in G$ such that $\alpha x = \beta_1$, $\alpha y = \gamma_1$ then neither of $\beta_2 x^{-1}$ and $\gamma y^{-1}$ are equal to $\alpha$ (otherwise $\beta_1 = \beta_2$ or $\gamma_1 = \gamma_2$) so there exists $h \in G_\alpha$ which maps one to the other $\beta_2 x^{-1} h = \gamma_2 y^{-1}$. Let $g = x^{-1} h y$ and compute $$(\beta_1,\beta_2) g = (\gamma_1,\gamma_2).$$
In the other direction if  $\Omega^2 \setminus \Delta$ is a single orbit the action is at least 2 (since we can fix any one element $\beta$ in the first component and map any other element $\gamma$ not equal to beta to any other element not equal to beta). So suppose the rank were larger than 2, then pick $\beta,\gamma$ in different $G_\alpha$ orbits of $\Omega \setminus \{\alpha\}$ and take $(\alpha,\beta), (\alpha,\gamma) \Omega^2 \setminus \Delta$ there's clearly no way to map from one to the other.

We can generalize this to $\Omega^t$ for any natural $t \le |\Omega|$. Again $\Delta = \{(\alpha,\alpha,\ldots)\}$ is a single orbit, but the interesting part is $\Omega^{(t)} = \{(\alpha_1,\alpha_2,\ldots)|\alpha_i \not = \alpha_j\}$.

Definition The action of $G$ on $\Omega$ is $t$-transitive if the induced action on $\Omega^{(t)}$ is transitive. This is equivalent to saying it can simultaneously map any $t$ distinct points to any other $t$ distinct points.

Lemma In terms of cosets, a group action is 2-transitive iff $G = G_\alpha \cup G_\alpha g G_\alpha$ for any $g \setminus G_\alpha$.
proof:  We saw previous that double cosets correspond to suborbits, write $G_\alpha = G_\alpha 1 G_\alpha$ to see this is the same as rank 2.

Lemma The action of $G$ on $\Omega$ is $t$-transitive iff the action of $G_\alpha$ on $\Omega \setminus \{\alpha\}$ is $(t-1)$-transitive.
proof: write this out.

Corollary If $G$ acts $t$-transitively on $\Omega$ and $|\Omega|=n$ then $|G|$ is divisible by $n(n-1)\cdots(n-t+1)$. WHICH THEOREM DOES THIS DEPEND ON? CHECK OTHER BOOK

Theorem In the natural action $S_n$ is $n$-transitive while $A_n$ is $n-2$ transitive and not $n-1$ transitive.
proof: This is obvious from the fact $S_n$ contains every permutation. For $A_n$ we use induction: it clearly holds for $A_3$. For $n \ge 3$ the stabilizer of any point of $A_n$ is $A_{n-1}$ so by the lemma we complete the induction.