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

  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 $|A|=|G.a|=[G:\mathrm{Stab}_G(a)]=|G|$.


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 $X$ :


  • 1
    $\begingroup$ I think you want to prove your last equality first, so that (using the transitivity of $G$ on $A$) you get that elements of the intersection are the ones you are considering in your first paragraph. The second part follows because conjugation in an abelian group is trivial (so all point stabilizers are trivial). See my answer. $\endgroup$ – David Hill Oct 30 '15 at 18:04

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