$k$-transitive implies $(k-1)$-transitive 
Let $G$ be a transitive permutation group on a set $X$. Suppose $k>1$
  and $a$ a fixed element of $X$. Then $G$ is $k$-transitive iff $G_a$
  is $(k-1)$-transitive on $X\backslash\{a\}$.

I am try to show $\Leftarrow$. Let $X^{[k]}$ be the set of all ordered $k$-tuples consisting of distinct elements of $X$. Suppose $(x_1,\ldots,x_ k),(y_1,\ldots,y_ k)\in X^{[k]}$. Then we have 4 cases: 


*

*$a\in\{x_1,\ldots,x_k\}$ and $a\in\{y_1,\ldots,y_k\}$;

*$a\in\{x_1,\ldots,x_k\}$ and $a\not\in\{y_1,\ldots,y_k\}$;

*$a\not\in\{x_1,\ldots,x_k\}$ and $a\in\{y_1,\ldots,y_k\}$;

*$a\not\in\{x_1,\ldots,x_k\}$ and $a\not\in\{y_1,\ldots,y_k\}$.


Assume (1) holds. Let $j,l\in[1,k]$ such that $a=x_j$ and $a=y_l$. Then
$$(x_1,\ldots,x_{j-1},x_{j+1},\ldots,x_ k),\,(y_1,\ldots,y_{l-1},y_{l+1},\ldots, y_ k)\in (X\backslash\{a\})^{[k-1]}.$$
Therefore there exists $\sigma\in G$ such that $\sigma(a)=a$ and
$$(y_1,\ldots,y_{l-1},y_{l+1},\ldots, y_ k)=(\sigma(x_1),\ldots,\sigma(x_{j-1}),\sigma(x_{j+1}),\ldots,\sigma(x_ k)).$$
I am not sure how to re-insert $x_j$ and $y_l$ back into these sequences if $j\ne l$. Am I doing something wrong? 
 A: A nice interpretation of transitivity to help understand the exercise is as follows: an action is transitive if there is a point which can be moved to any other point (or conversely, if any other point can be moved to it). For example, the real numbers acting on the real number line by translation is transitive because you can always move anything to $0$, or move $0$ to anything.
In this case, $G$ is $k$-transitive on $X$ means the action on $X^{[k]}$ (ordered lists of $k$ distinct elements of $X$) is transitive. Suppose you know $G$ is $(k-1)$-transitive of $X\setminus\{a\}$. Then automatically you know it's possible to go between any two lists of the form $(x_1,\cdots,x_{k-1},a)$. Then the question becomes, can an arbitrary list like $(u_1,\cdots,u_k)$ be turned into a list for the form $(x_1,\cdots,x_{k-1},a)$? Hint: yes, it's possible, since all you need to do is turn $u_k$ into $a$, and we know $G$ is at least transitive on $X$...
So to go from $(x_1,\cdots,x_k)$ to $(y_1,\cdots,y_k)$ we can go
$$ (x_1,\cdots,x_k)\mapsto(u_1,\cdots,u_{k-1},a)\mapsto(v_1,\cdots,v_{k-1},a)\mapsto(y_1,\cdots,y_k). $$
The first comes from using a group element to go $x_k\mapsto a$, the last from an element to go $y_k\mapsto a$ (and then inverting it), and then at the end you can pick a way to go from $(\vec{u},a)$ to $(\vec{v},a)$.
A: You can do it without subdividing into four cases.
In brief: choose $\sigma_1,\sigma_2\in G$ with $\sigma_1(x_1)=a$, $\sigma_2(a)=y_1$, then choose $\sigma_3 \in G_a$ with $\sigma_3(\sigma_1(x_i)) = \sigma_2^{-1}(y_i)$, for $2 \le i \le k$. Then $\sigma_2\sigma_3\sigma_1(x_i)=y_i$ for $1 \le i \le k$.
