# How to prove the equivalence of definitions of doubly transitive group

Given a group $G$ which acts on a set $X$ we define that a group $G$ acts doubly transitive on $X$ as $G$ is transitive on $X$ and $G_x$ is transtive on $X \backslash \{x\}$ where $G_x$ the stabilizer is and $x \in X$. I want to prove that for $x\neq x'$ and $y \neq y'$ an element $g \in G$ exists with $(g(x),g(x')) = (y,y')$. How can I do that ?

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Let's just think about our definitions. We know that there is a $g_1 \in G$ with $$g_1 \cdot x = y.$$ We also know that for any point $z \ne x$, there is a $g_2 \in G_x$ with $$g_2 \cdot x' = z.$$ Let's check that we can take $$z = g_1^{-1}\cdot y';$$ in other words, we need to check that $g_1^{-1} \cdot y' \ne x$. If this were the case, then we would have $$g_1\cdot x = y',$$ but we already know that $g_1 \cdot x = y$ and that $y \ne y'$. Now I claim that $g_1g_2$ is the required element. We have $$(g_1g_2)\cdot x = g_1\cdot (g_2 \cdot x) = g_1 \cdot x = y,$$ since $g_2 \in G_x$, and $$(g_1g_2)\cdot x' = g_1\cdot (g_2 \cdot x') = g_1 \cdot (g_1^{-1} \cdot y') = y'.$$ So intuitively, what we're doing here is first acting by an element of the stabilizer so that $x'$ has moved to the right position for it to then be moved to $y'$ when we act by the element moving $x$ to $y$.

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Greath thanks to you. Now i can sleep :) – Epsilon Sep 30 '12 at 22:56
Is the converse of this theorem true ? – Epsilon Oct 1 '12 at 15:17
Yes, it is. Perhaps you would like to try showing it yourself. – NKS Oct 1 '12 at 21:50

The idea will be to pick $g_1 \in G$ with $g_1 \cdot x = y$ and then use $g_2 \in G_x$ to move $x'$ to $y'$. We'll have to use $g = g_1 g_2$ to insure that $g \cdot x = y$. And in order to have $g \cdot x' = y'$, we have to pick $g_2 \cdot x' = g_1^{-1} \cdot y'$.

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A small comment, but it wasn't clear to me at first why such a $g_2$ had to exist (as in, why couldn't $g_1^{-1}\cdot y' = x$?) It turns out to be easy, but I do think it is worth checking, or at least noting. – NKS Sep 30 '12 at 22:46
You're absolutely right, +1 to your thorough response. – Michael Joyce Sep 30 '12 at 22:48
You seem to be assuming that the $x$ in the second sentence of the problem is the same as the $x$ in the first sentence, but it need not be. You only know that $G_x$ is transitive on $\Omega \setminus \{x\}$ for some specific $x \in X$, but you need to prove the condition for all $x,x',y,y'$. – Derek Holt Oct 1 '12 at 15:19
@DerekHolt: Although it's not clearly stated in the problem, my assumption was that you know $G_x$ is transitive on $X \setminus \{x\}$ for all $x \in X$. Can one really prove the result only assuming it holds for a single $x \in X$? – Michael Joyce Oct 1 '12 at 16:12
@Michael Joyce. Yes one can, but it needs an extra step. Suppose $G_x$ is transitive on $X \setminus \{x\}$ for some fixed $x$ and $y \in X$. Let $y_1,y_2 \in Y \setminus \{y\}$. Since $G$ is transitive on $X$, there exists $g_1 \in G$ with $g_1.y=x$. So $g_1.y_1,g_1.y_2 \in X \setminus \{x\}$ and there exists $g_2 \in G_x$ with $g_2.g_1.y_1=g_1.y_2$. Now $g = g_1^{-1}g_2g_1$ satisfies $g \in G_y$ with $g.y_1=y_2$, and we have the condition for all elements of $X$. – Derek Holt Oct 1 '12 at 17:49