On Group action and blocks of subgroups of the symmetric Group

Question

this exercise is from Dummit and foote , page 117 , # 7.d

prove : a transitive group $G$ is primitive on $A$ iff for each $a \in A$ , the only subgroups of $G$ containing $G_a$ are $G_a$ and $G$ where , $G_a$ is the stabilizer of $a$ $G$ is subgroup of $S_n$ - symmetric group -

my attempt ,

i proved one side , i proved that if the only subgroups of $G$ containing $G_a$ are $G_a$ and $G$ then , the transitive Group $G$ is primitive

but i couldn't prove the other side the other side .

what i had thought in that if i can prove that , if $H$ is subgroup of $G$ that contains Ga then H = GB for some subset $B$ of $A$ , $a$ is in $B$ where $GB$ = { $g \in G$ : $g$ . $B = B$, . is group action operation }

then , I can prove the statement

but I don't know excatly how to do this ! I don't know if this is true to prove it !!

i hope that you give me some hints which able me to create my own solution

thanx !

Answer 1

Suppose $G$ is primitive on $A$. Hence the only blocks of $A$ are the singleton sets and $A$ itself. Take $a\in A$ and suppose $H\leq G$ is a subgroup such that $G_a\leq H\leq G$. Consider $Ha=\{ha:h\in H\}$.

• Prove that $Ha$ is a block.

Let $\sigma\in G$ as in the text. Suppose there exists some $x\in Ha\cap\sigma(Ha)$. So there are $h_1,h_2\in H$ such that $h_1a=\sigma h_2a$. This implies $h_1^{-1}\sigma h_2a=a$, hence $h_1^{-1}\sigma h_2\in G_a$. But $G_a\leq H$, and thus $\sigma\in H$ by left and right multiplying by $h_1$ and $h_2^{-1}$. But then $\sigma(Ha)=Ha$ since $Ha$ is invariant under the action of $H$, so $Ha$ is a block.

After proving that, we know there are two cases by primitivity, so either $Ha=A$, or $Ha=\{a\}$.

If $Ha=A$, for every $g\in G$ we can find $h\in H$ such that $ga=ha$, since the action is transitive. But then $h^{-1}ga=a$, thus $h^{-1}g\in G_a\leq H$, whence $g\in H$. It follows immediately that $H=G$ in fact.

The other case is simpler.

If $Ha=\{a\}$, then for every $h\in H$, $ha=a$, thus $H\leq G_a$, so $H=G_a$.

So if $a\in A$, and $H\leq G$ with $G_a\leq H\leq G$, we have either $H=G_a$ or $H=G$. So $G_a$ is maximal in $G$, as we needed to show.