Stabilizers, Orbits, and the Symmetric Group

Question

I have a homework question I am a bit confused on.

Let $S_n$ act on the set of ordered pairs $(i,j)$, $1\leq i,j \leq n$ such that $\omega \cdot (i,j)= (\omega i,\omega j)$.

I have already proved that there are two orbits. One containing all ordered pairs $i=j$ and the other all other ordered pairs.

I am having trouble finding the stabilizer of these orbits considering they are both infinite and just generally struggling with deciphering how to find stabilizers.

Any Help is greatly appreciated.Thanks!

Answer 1

They are not infinite at all, there are only finitely many choices for $(i,j)$ namely $n^2$, $n$ choices for the first entry and $n$ for the second.

So for the orbit stabilizer theorem, you know that if $C_x$ is the stabilizer of $x$ then

$$|\text{Orb}_G(x)||C_x|=|S_n|=n!$$

So for the diagonal orbit, i.e. the orbit of $(i,i)$ we know that since the orbit is all other diagonal elements, the stabilizer size must have size $n!/n=(n-1)!$ which suggests the stabilizer is a copy of $S_{n-1}$. Indeed we see that the subgroup of $S_n$ which uses just the symbols which are not $i$ forms a copy of $S_{n-1}$ which stabilizes $(i,i)$ hence must be the entire stabilizer by noting the cardinalities match.

Similarly we see that the remaining elements are all in the same orbit, which has size $n^2-n=n(n-1)$, so $n!/n(n-1)=(n-2)!$.

This suggests the stabilizer might be a copy of $S_{n-2}$ and indeed it is, consider the copy of $S_{n-2}$ which uses all symbols except $i$ and $j$, this has the right order and it clearly stabilizes $(i,j)$ so by the theorem it must be the entire stabilizer.

Answer 2

Just follow the definition of stabilizer. Since $G$ has 2 orbits then we have two different classes of stabilizers.

1- $\alpha = (i,i)$.

In this case we have $$St_{S_n}(\alpha) = \{g \in S_n | (gi,gi)=(i,i),$$ $$=\{g \in S_n | gi =i \}$$ So $St_{S_n}(\alpha)$ is a subgroup of $S_n$ that fixes $i$. You should note that this subgroup is isomorphic to $S_{n-1}$.

2- $\beta = (i,j)$, where $i \ne j$.

In this case we have $$St_{S_n}(\beta) = \{g \in S_n | (gi,gj)=(i,j),$$ $$=\{g \in S_n | gi =i, gj=j \}$$

So $St_{S_n}(\beta)$ is a subgroup of $S_n$ that fixes $i$ and $j$. This subgroup is isomorphic to $S_{n-2}$.