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 ?
How to prove the equivalence of definitions of doubly transitive group
2 Answers
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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0Yes, it is. Perhaps you would like to try showing it yourself. – 2012-10-01
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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0Greath thanks for this ! My intention was that $G_x$ is transitive on $X\backslash \{x\}$ for all $x \in X$. But this is even greater :) – 2012-10-01