Understanding Generalised Quadrangles I have a project to do on Generalised Quadrangles, specifically GQ(2,2). The project needs to have information about the construction of GQ(2,2), to prove this construction meets the conditions of a generalised quadrangle and showing that $S_{6}$ acts "flag transitively" on this object.
I'm aware now of the definition of $GQ(2,2)$, and I have some idea on how to construct it. Essentially, it seems you take $6$ points, and then take all possible triplets of pairs that can be formed from this (and there are $\binom{6}{2} = 15$ of these unordered pairs) and then you just put lines between triplets who share points. However, I'm struggling to prove that this matches the definition of a generalized quadrangle. Furthermore, I don't really understand what flag transitivity means, or how to show that $S_{6}$ acts in this way. 
 A: First I'll go over flag transitivity in general, flesh out your construction of $GQ(2,2)$, and then give a hint for how to prove that the action of $S_6$ on $GQ(2,2)$ is flag-transitive.
First, the definition:

Let $Q$ be a generalized quadrangle, and $G$ be a group acting on $Q$. We say that the action of $G$ on $Q$ is flag transitive if for every pair $((p_1,l_1), (p_2,l_2))$ where $p_i$ is a point incident with the line $l_i$, there exists some $g\in G$ such that $(p_1,l_1)^g = (p_2,l_2)$.

If you're familiar with point transitivity and/or line transitivity, this should feel a little similar. A flag is just an incident point-line pair, and being flag transitive means that the action takes one flag to any other flag. Note that this is stronger than both point and line transitivity.
Now, for your construction. I'm going to go over it in detail mostly for myself; if I am misunderstanding your idea, let me know!
Take $S = \{1, 2, 3, 4, 5, 6\}$ and consider the set $T = \{\{a,b\}: a,b\in S, a\ne b\}$. As you mentioned, since ${6\choose 2} = 15$, the set $T$ has $15$ elements. Take $T$ to be the point-set of $GQ(2,2)$. Then three points $A,B,C\in T$ are all pairwise collinear (equivalently, define a unique line) if and only if $A\cup B\cup C= S$. Thus, take the set $R = \{\{A,B,C\}:A,B,C\in T, (A\cup B\cup C) = S\}$ to be the set of lines of $GQ(2,2)$. It should be shown that $R$ also has $15$ elements (since $GQ(s,t)$ has $(t+1)(st+1)$ lines). To prove that this yields your generalized quadrangle, you should prove three things:


*

*For every $A\in T$, there exists exactly three lines incident with $A$. This means you should find exactly three pairs $\{B_i,C_i\}$ where $\{A,B_i,C_i\}\in R$.


*For every $L\in R$, there exists exactly three points incident with $L$. This comes more-or-less from the definition (unless I'm missing a subtlety).


*For every pair $(A,L)$ where $A$ and $L$ are not incident, there exists a unique point $B$ incident with $L$ and collinear with $A$.

If these all hold, then (since $GQ(2,2)$ is unique), you will have truly constructed $GQ(2,2)$.
To prove that the action of $S_6$ on $GQ(2,2)$ is flag transitive, you should ask yourself first: what is the action? If $\sigma\in S_6$, then for $\{a,b\}\in T$, define $\{a,b\}^{\sigma}:= \{\sigma(a), \sigma(b)\}$. This gives a natural action on the points. Similarly, for $\{A,B,C\}\in R$, define $\{A,B,C\}^{\sigma}:= \{A^{\sigma}, B^{\sigma}, C^{\sigma}\}$. Lastly, for $A\in T$ and $L\in R$ where $A$ is incident to $L$, define $(A,L)^{\sigma}:= (A^{\sigma}, L^{\sigma})$. You should prove that these definitions are indeed group actions!
Hopefully, this gives a good foundation for how to show that this action of $S_6$ on flags is in fact transitive. You might want to show that for any flag $(A,L)$, there exists some $\sigma\in S_6$ such that $(A,L)^{\sigma} = (\{1,2\}, \{\{1,2\},\{3,4\},\{5,6\}\})$. This almost immediately implies transitivity (why?).
