doubly transitive groups are primitive

Theorem.

Every doubly transitive group is primitive (http://planetmath.org/PrimativeTransitivePermutationGroupOnASet).

Proof.

Let $G$ acting on $X$ be doubly transitive. To show the action is , we must show that all blocks are trivial blocks; to do this, it suffices to show that any block containing more than one element is all of $X$. So choose a block $Y$ with two distinct elements $y_1,y_2$. Given an arbitrary $x\in X$, since $G$ is doubly transitive, we can choose $\sigma \in G$ such that

$$\sigma \cdot (y_1,y_2)=(y_1,x)$$

But then $\sigma \cdot Y\cap Y\ne \mathrm{\varnothing }$, since $y_1$ is in both. Thus $\sigma \cdot Y=Y$, so $x\in Y$ as well. So $Y=X$ and we are done. ∎

Title	doubly transitive groups are primitive
Canonical name	DoublyTransitiveGroupsArePrimitive
Date of creation	2013-03-22 17:21:50
Last modified on	2013-03-22 17:21:50
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	7
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 20B15