Blocks Primitivity and Multiple Transitivity Examples

If $H\le C_n$, then every translate $a+H$ is either $H$ itself or a disjoint coset of $H$. So each coset of $H$ is a block, and the cosets of $H$ form a block system.

Conversely, let $B$ be a block containing $0$. For any $x\in B$, the translate $x+B$ meets $B$ at $x$, so [L1] gives $x+B=B$. Hence $B$ is closed under subtraction: if $x,y\in B$, then $y\in x+B$ implies $y-x\in B$. Therefore $B$ is a subgroup of $C_n$.

Every block is a translate of one containing $0$, so steps 1.1-1.2 show that the blocks are exactly the cosets of subgroups. When $n$ is composite, $C_n$ has a proper nontrivial subgroup; when $n$ is prime, it does not.

For the action of $S_n$, a permutation is determined uniquely by the images of the ordered tuple $(1,\dots ,n)$, and every ordered $n$-tuple of distinct points is another listing of $\{1,\dots ,n\}$. So this action is sharply $n$-transitive by [L1].

For the action of $A_n$ with $n\ge 4$, take two ordered $(n-2)$-tuples of distinct points and let $c_1,c_2$ and $d_1,d_2$ be the two complementary points. Some $\sigma \in S_n$ sends the first full $n$-tuple $({\alpha }_1,\dots ,{\alpha }_{n-2},c_1,c_2)$ to the second $({\beta }_1,\dots ,{\beta }_{n-2},d_1,d_2)$.

If $\sigma \in A_n$, then $\sigma$ already sends ${\alpha }_i$ to ${\beta }_i$ for $1\le i\le n-2$. If $\sigma \mathrm{\notin }{A}_n$, compose it with the transposition $(d_1{d}_2)$, which fixes each ${\beta }_i$ and reverses parity. So in either case there is an even permutation sending the first $(n-2)$-tuple to the second. Hence the natural action of $A_n$ is $(n-2)$-transitive.

Let $(x_1,x_2)$ and $(y_1,y_2)$ be ordered pairs of distinct points of $V$. Then $x_2-x_1$ and $y_2-y_1$ are nonzero vectors, so there is some invertible linear map $A$ with $A(x_2-x_1)=y_2-y_1$.

Define $b:=y_1-A{x}_1$ and $g(x):=Ax+b$. Then $g(x_1)=y_1$ and $g(x_2)=A{x}_2+b=A(x_2-x_1)+y_1=y_2$. So $\mathrm{AGL}(V)$ is doubly transitive by [L1].

By [L2], the action is primitive.

The action of $D_n$ on the $n$ vertices is transitive, so if $n$ is prime, [L1] makes it primitive.

Suppose $n$ is composite and let $d$ satisfy $1<d<n$ and $d\mid n$. Put $B:=\{0,d,2d,\dots ,n-d\}\subseteq \mathbb{Z}\mathrm{/}n\mathbb{Z}$. Rotations send $B$ to its residue-class translates modulo $d$, and reflections send residue classes modulo $d$ to residue classes modulo $d$ as well. Hence every dihedral image of $B$ is either $B$ itself or a disjoint residue class, so [L2] makes $B$ a block.

Because $1<d<n$, the block $B$ is neither a singleton nor all of $\mathbb{Z}\mathrm{/}n\mathbb{Z}$. So composite degree produces nontrivial blocks.

The matrices $\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right)$ give translations $x\mapsto x+b$, so the stabilizer of $\mathrm{\infty }$ is transitive on ${\mathbb{F}}_q$. The matrix $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ sends $\mathrm{\infty }$ to $0$; composing it with translations sends $\mathrm{\infty }$ to any finite point. All these matrices have determinant one, so they act through $G$, not merely through the full projective linear group. Hence the action is transitive, and its point stabilizer is transitive on the other points: it is doubly transitive and therefore primitive by [L1]. A fractional linear map fixing $\mathrm{\infty }$, $0$, and $1$ is scalar as a matrix, hence the projective action is faithful.

The subgroup $A$ fixes $\mathrm{\infty }$, is abelian under composition, and is normal in the stabilizer of $\mathrm{\infty }$ because conjugating a translation by an affine map gives another translation.

Conjugating $A$ by the inversion $x\mapsto -1\mathrm{/}x$ gives the lower-unitriangular subgroup. The upper and lower unitriangular subgroups generate ${\mathrm{SL}}_2({\mathbb{F}}_q)$ by Gaussian elimination, so their projective images generate $G$. Thus the conjugates of $A$ generate $G$, and [L2] applies.

For $(f,k)\in H{\wr }_{\mathrm{\Sigma }}K$, the image of the fiber $F_{\sigma }$ is exactly $F_{k\cdot \sigma }$ by [L1]. Hence every group element sends a fiber to a fiber.

Each fiber is nonempty because $B$ is nonempty, and two distinct fibers are disjoint, so [L2] makes each $F_{\sigma }$ a block. The collection of all fibers partitions $B\times \mathrm{\Sigma }$.

If $\mathrm{\mid }B\mathrm{\mid }>1$ and $\mathrm{\mid }\mathrm{\Sigma }\mathrm{\mid }>1$, then each fiber is a proper non-singleton subset. So the fiber partition is a nontrivial block system.

The nonzero squares in ${\mathbb{F}}_7$ are $\{1,2,4\}$, and every nonzero element is either a square or the negative of a square. So for any distinct $x,y,u,v\in {\mathbb{F}}_7$, after possibly swapping $u$ and $v$ there is $a\in \{1,2,4\}$ with $a(y-x)=v-u$. Then the affine map $z\mapsto a(z-x)+u$ lies in $G$ and sends $\{x,y\}$ to $\{u,v\}$. Thus the action is $2$-homogeneous.

Every element of $G$ multiplies differences by a square: if $g(z)=az+b$, then $g(y)-g(x)=a(y-x)$ with $a\in \{1,2,4\}$. Since $1$ is a square and $3$ is not, no element of $G$ sends the ordered pair $(0,1)$ to $(0,3)$. So the action is not $2$-transitive.

Step 1.1 gives $2$-homogeneity while step 1.2 denies $2$-transitivity, refuting the statement.

The regular action of $C_n$ on itself is transitive, since every point is reached by a translation.

When $n$ is composite, [L1] gives a nontrivial block system coming from a proper nontrivial subgroup of $C_n$. So this transitive action is not primitive.

Steps 1.1 and 1.2 refute the statement.

In the action of $D_6$ on $\mathbb{Z}\mathrm{/}6\mathbb{Z}$, the subset $B=\{0,2,4\}$ is a block by [L1].

The reflection $x\mapsto 1-x$ sends $B$ to $\{1,3,5\}$, which is disjoint from $B$ and not equal to it. So $B$ is a block but not an invariant subset.

This refutes the statement.

By [L1], the square-affine action satisfies the hypothesis of the statement and fails its conclusion.

Therefore the statement is false.

By [L2], $A_n$ is normal in $S_n$, and it is nontrivial because it contains the $3$-cycle $(123)$. The natural $S_n$-action is faithful, and [L1] makes it doubly transitive, hence primitive by [L3].

The element $(123)\in A_n$ fixes the point $4$, so the action of $A_n$ is not free and therefore not regular. Thus $A_n$ is a nontrivial normal subgroup of a faithful primitive action that is not regular.

The action of $C_4$ on itself is transitive, but [L2] shows it is not primitive.

The subgroup $2C_4=\{0,2\}$ is nontrivial, normal, and not transitive: its orbits are $\{0,2\}$ and $\{1,3\}$. So the transitivity conclusion singled out in [L1] fails once primitivity is removed.

The regular action is faithful and transitive, while step 2.1 gives a nontrivial normal subgroup that is not transitive. Therefore the statement is false.