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For the degree-ten partition action of S6, take unordered partitions of six letters into two triples. There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$. Fix $A\mid A^c$. Its stabilizer is $(S_3\times S_3)⋊S_2=S_3\wr S_2$, permuting each triple and interchanging them. The subgroup generated by a 3-cycle in each triple has order nine, so is a Sylow subgroup of $S_6$. Any normalizer preserves its two three-point orbits; conversely all the indicated within-triple permutations and the swap normalize it. This proves the asserted normalizer exactly.

For any different partition $C\mid C^c$, $|A\cap C|$ is one or two after excluding the identical partition. Interchanging $C$ and $C^c$ identifies these two possibilities. The stabilizer of $A\mid A^c$ is transitive on each fixed intersection pattern, hence on all other nine partitions. The degree-ten action is two-transitive. It is also faithful: two distinct partitions have a common refinement with two singleton cells and two cells of size two. Every unordered pair of letters occurs as those singleton cells. A permutation fixing every partition therefore fixes every pair and consequently every letter.

For the subgroup factorization from coset transitivity, use cosets $Hg$ (right cosets) and right multiplication by $K$. The orbit of $H$ is precisely $\{Hk:k\in K\}$. This is every coset if and only if every $g$ can be written $hk$, provingA natural $S_5$ fixing one letter is transitive on the ten partitions: the triple containing the fixed letter is determined by any two of the other five. Taking $H=S_3\wr S_2$, $K=S_5$ gives $S_6=HK$; inversion gives the requested order as well,

Now let $B$ be the binary nondegenerate alternating bilinear form. Its binary quadratic refinements form an affine space: the difference of two such forms is linear, so for a fixed $Q_0$ every one is uniquelyIn a symplectic basis each refinement isA two-dimensional summand with $(a_i,b_i)\ne (1,1)$ admits a basis of singular vectors paired to one and is hyperbolic; the $(1,1)$ summand is anisotropic. Two planes spanned by anisotropic vectors for a form together become two hyperbolic planes of a quadratic form: with anisotropic basis pairs $(e_1,f_1),(e_2,f_2)$, the singular vectors $e_1+e_2,f_1+e_2$ pair to one; their orthogonal complement for a sesquilinear form has singular paired basis $e_1+f_1+f_2,e_1+f_1+e_2+f_2$. Consequently every refinement is equivalent to either $m$ hyperbolic planes of a quadratic form or $(m-1)$ hyperbolic planes of a quadratic form and one anisotropic plane (all three nonzero vectors are anisotropic vectors for a form).

These two types are different. Indeed the character sum of each plane is $+2$ or $-2$, soThus there are exactly two symplectic orbits $Q^{+}(B),Q^{-}(B)$, the signs denoting $\mathrm{Arf}=0,1$. Counting the three coefficient pairs with product zero and the one with product one givesThe same character sum gives the corresponding numbers of singular vectors, including zero, in a fixed form of either type.

Fix a form $Q$ of type $ϵ$. Completing the square by $x\mapsto x+v$ showsThe stabilizer of $Q$ is exactly $O(V,Q)$. It sends the parameter $v$ by its natural linear action or inverse action according to the chosen left/right convention. Other forms of the same type correspond to nonzero singular vectors $v$, and all forms of opposite type correspond to nonsingular vectors $v$. For two nonzero vectors of the same $Q$ value, the map between their spans is an isometry of a space with a form. Witt's lemma extends it, so the orthogonal stabilizer is transitive on each of these two vector sets. This proves the two-transitive binary quadratic-form actions: the symplectic group over a finite field is transitive on each type and its point stabilizer is transitive on all other members of that type.

In particular $O_{2m}^{-}(2)$ is transitive on the plus orbit. Identify that orbit with the cosets (right cosets) of a plus stabilizer, using $Q^g(x)=Q(gx)$ as a right action. The coset criterion then provesBoth genuine two-transitive actions require $m\ge 2$. At $m=1$ the plus orbit has three forms and the minus orbit is a singleton; transitivity and the factorization remain true, while two-transitive action on the singleton is only vacuous if one's convention permits degree one. No special simplicity assertion is needed for this argument. At $m=2$ the orbit sizes are ten and six, linking these actions with $S{p}_4(2)\cong S_6$ and the partition action above.