Prove $G$ acts $(k + 1)$-transitively on $A$ $\iff$ $\exists a \in A, G_{(\{a\})}$ acts $k$-transitively on $A \setminus\{ a \}$

Question

Definitions

The pointwize stabiliser $G_{(A)}$ is defined as the set $\{g \in G | \forall a \in A: g(a) = a \}$

A subgroup $G ≤ Sym(X)$ acts $k$-transitively on a subset $A ⊆ X$ if $|A| ≥ k$ and $\forall a_1, . . . , a_k, b_1, . . . , b_k ∈ A$ such that $a_i \ne a_j$ and $b_i \ne b_j, \forall i \ne j$ there exists $g ∈ G$ such that $g(a_i) = b_i$ for all $i$ between $1$ and $k$

Problem

Let $G \le Sym(X)$ and $A \subseteq X, |A| \ge k+1$ and $G$ acts $1$-transitively on $A$

Prove the following are equivalent:

1. $G$ acts $(k + 1)$-transitively on $A$
2. $\forall a \in A, G_{(\{a\})}$ acts $k$-transitively on $A \setminus\{ a \}$
3. $\exists a \in A, G_{(\{a\})}$ acts $k$-transitively on $A \setminus\{ a \}$

I've been able to prove 1 $\iff$ 2 $\implies$ 3 so all I need left is 3 $\implies$ 1 or 2

Answer 1

I show $(3) \Rightarrow (2)$.

Let $x \neq a$, and let $a_1,\ldots,a_k$ be pairwise distinct elements of $A \backslash \{x\}$, $b_1,\ldots,b_k$ be also pairwise distinct elements of $A \backslash \{x\}$.

Let $\sigma \in G$ mapping $a$ to $x$. Let $a’_i=\sigma^{-1}(a_i)$, $b’_i=\sigma^{-1}(b_i)$. They’re elements of $A \backslash \{a\}$, the $a’_i$ are pairwise disjoint, the $b’_i$ are pairwise disjoint, so there is a $\tau \in G_{(\{a\})}$ such mapping $a’_i$ to $b’_i$. Then $\tau’=\sigma \circ \tau \circ \sigma^{-1}$ maps $a_i$ to $b_i$ and you easily check $\tau’ \in G_{(\{x\})}$.

Answer 2

$3\to1$: Let $a_1,...,a_{k+1},b_1,...b_{k+1}\in A$ (with the appropriate conditions). I write $\bar{a}=(a_1,...a_{k+1})$

1. If $a_1=a$ then since $G$ acts $1-$ transitively there exists some $t\in G:\ tb_1=a$. So $t\overline{b}=(a,b_2,...,b_{k+1})$. There is some $h\in G_{\{a\}}:\ h(a_2,...,a_{k+1})=(b_2,...b_{k+1})$ so $h(a_1,a_2,...,a_{k+1})=(a,b_2,...b_{k+1})=t\overline{b} \Rightarrow t^{-1}h\bar{a}=\bar{b}$ $\checkmark$
2. If $a_1\not=a $ then since $G$ acts $1-$transitively $\exists g\in G$ st $ga_1=a$ hence $g(a_1,...,a_{k+1})=(a,a_2,...,a_{k+1})$ and we are in case $1$.