Group action of Symmetric group on arbitrary set A

Question

I was recently given this problem in my Abstract Algebra course dealing with group actions, stating the following:

Let A be a non-empty set and $ S_A $ is its symmetric group. Now assume we have a subgroup $ G < S_A $ which acts transitively on A.

(a). Let $ a \in A $ and denote $ H = Stab_G(a) $. We are to show that $ \bigcap_{g \in G} gHg^{-1} = \{1\} $.

(b). We are to show that if G is also Abelian then we have $ |G| = |A| $ (equal cardinalities not necessarily finite).

Now I have tried to do part a as follows: I know the stabilizer is a subgroup of G and so are all its intersections so the intersection I have here is also a subgroup of G so it has the identity element, so a proof by contradiction would involve assuming the intersection contains an element of g which we may call $ k $ which is not the identity and after some manipulation we see that for all $ g \in G $ we must have $ g \cdot a = k \cdot (g \cdot a ) $ and I have no idea how to reach a contradiction given transitivity of the action of G on A. As for part b, now idea how to show a bijection and how to use Abelian group I truly need the help on this. Thanks all

Answer 1

1. First prove the following exercise: $$gHg^{-1}=g\mathrm{Stab}_G(a)g^{-1}=\mathrm{Stab}_G(g.a)$$ Since the action of $G$ is transitive, every $b\in A$ is equal to $g.a$ for some $g\in G$, so we have $$\cap_{g\in G}gHg^{-1}=\cap_{a\in A}\mathrm{Stab}(a)=:K$$ Now, $K$ is the set of elements $g\in G$ such that $g.a=a$ for all $a\in A$. But $G\leq S_A$, and the only element in $S_A$ with this property is $1_A$.

2. Now, if $G$ is abelian, then $gHg^{-1}=H$ for all $g\in G$. In particular, $\{1\}=\cap_g gHg^{-1}=\cap_g H=H$. So $\mathrm{Stab}_G(a)=\{1\}$ for any $a\in A$. Since the action of $G$ is transitive, $A=G.a$ for (any) $a\in A$, and so $|A|=|G.a|=[G:\mathrm{Stab}_G(a)]=|G|$.

Answer 2

For question 1. Take $f$ an element of $G$ acting trivially. Then for all $a$ in $A$, $f.a=a$. Now $f.a:=f(a)$ so that $f(a)=a$. This is true for all $a$ so $f$ is the identity function on $A$. So $f$ is indeed the neutral element of $S_A$.

For the second question demonstrate then use the following result : When $G$ acts on a set $X$ then for all $g$ in $G$ and $x$ in $X$ :

$$Stab(g.x)=gStab(x)g^{-1}$$