Permutation group action on the set of $n$-sets of a $v$-set

Question

I am trying to understand a proof and I have the following questions.

Let $X$ be the set of $n$-sets of a $v$-set i.e. $|X| = \binom{v}{n}$ and $G=S_V$ the Permutation group. Then $G$ acts transitively on $X$ via natural action $g\{a_1,...,a_n\} = \{ga_1,...,ga_n\}$.

1. I understand that $G=S_V$ acts transitively on $\{1,...,v\}$ but I dont see why it is also true for a set $\{1,...,n\}$ (with for example $n \leq \frac{v}{2}$. Do you just take the same natural action for the first $n$ numbers and let it be constant for the other $\{n+1,...,v\}$ numbers i.e. $\sigma \in S_V, \sigma(k) = k \quad \forall k\in \{n+1,...,v\}$ ?

Then it is said that for all $x,y \in X$ there $ \exists g\in G: y=gx , x=gy$. So that $g$ is a sequence of disjoint transpositions.

2. I dont see why the first statement is true at all to be honest. I am probably completely wrong here but I tried the following. If we let $v=4$ then we have $X = \{\{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}\}$. Let $x=\{1,2\}, y= \{2,3\}$ then there must be a $g\in G$ so that $$g\{1,2\} = \{g(1), g(2)\} = \{2,3\}$$ $$g\{2,3\} = \{g(2), g(3)\} = \{1,2\}$$ so $g(2) = 3$ in the first equation but $g(2) = 1$ in the second.

Furthermore i have one more question.

3. If I know that a finite Group $G$ acts transitively on a finite set $X$ does this also imply the Group acts transitively on $X \times X$ ?

Many thanks

Answer 1

(1) The group $G=S_v$ of permutations of $\{1,\cdots,v\}$ does not act on the proper subset $\{1,\cdots,n\}$ when $n<v$, and no one said it does, so you're arguing with something nobody said. The claim is that $S_v$ acts on the set $X$ of all size-$n$ subsets of $\{1,\cdots,v\}$. Observe $X$ and $\{1,\cdots,n\}$ are very different sets. You seem to understand this fine in part $(2)$, so I'm not sure why you have a different understanding in part $(1)$?

(2) Suppose we have two equal-size subsets $A,B\subseteq\{1,\cdots,v\}$. Draw a Venn diagram that represents $A$ and $B$; in general it will also have numbers in the middle. Because $A$ and $B$ have the same size, there are as many numbers strictly on the left circle (not in the middle) as there are strictly on the right circle (not in the middle). If you draw out a one-to-one correspondence between these numbers, you pair them up, and if you interpret each pair as a transposition and let $g$ be their product, then $B=gA$ and $A=gB$.

(3) In general if any group $G$ acts on any set $X$ with more than one element, $G$ cannot act transitively on the cartesian product $X\times X$. In other words, there cannot be just one orbit. There are at least two orbits, because the diagonal $\Delta=\{(x,x)\mid x\in X\}\subsetneq X\times X$ is an orbit and a proper subset. The complement of the diagonal could be one or more orbits; if it is one orbit we say the action $G\curvearrowright X$ is doubly transitive.

In particular if $G=S_v$ acts on $X$, the collection of size-$n$ subsets of $\{1,\cdots,v\}$, and $v>3$, then the action $G\curvearrowright X$ is not doubly transitive because no element of $G$ can turn a pair of disjoint size-$n$ subsets into a pair of non-disjoint size-$n$ subsets. Indeed, the orbits of $G\curvearrowright X\times X$ are parametrized by intersection size; that is, for any two pairs $(A,B),(C,D)\in X\times X$ (where $A,B,C,D$ are size-$n$ subsets of $\{1,\cdots,v\}$), there exists a $g\in G(=S_v)$ for which $(C,D)=g(A,B)$ if and only if $|A\cap B|=|C\cap D|$. You can construct $g$ using a Venn diagram again, but it generally won't be a product of disjoint transpositions.

Answer 2

1. You seemed to know the answer to you first question in your second question, so I'm not sure where the confusion lies. To be clear, if $V = \{1,2,3\}$ and $X = \{\{1,2\},\{1,3\},\{2,3\}\}$ is the set of $2$-element subsets of $V$, then $G$ acts naturally on $X$ by "entering into the innermost bracket", exactly as you did in your second question.

2. Order doesn't matter in sets, so the first line does not imply $g(2) = 3$, and the second line does not imply $g(2) = 1$. In fact, you can see that if $g(2) = 2$, $g(1)=3$, and $g(3)=1$, then the statement holds (at least for this example, but it should also make clear how it always holds).

3. Nope, only very trivial sets and groups satisfy this, and in general it is not the case (now that you know this fact, try to find an example where your question doesn't hold).