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Let $G$ be a finite group acting on a finite set $X$. If the action is primitive then the stabilizers are maximal subgroups of $G$ (converse also true). Is there any criteria to get maximal subgroups as stabilizers, with some restriction on the "action"?

For ex. the action of $S_3$ on $X=\{1,2,3\}$ is primitive, and so stabilizers are maximal subgroups of $G$, which are Sylow-2 subgroups of $S_3$. But how can we obtain the maximal subgroup $\langle (123)\rangle$?

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You want to use \langle instead of < and \rangle instead of >. – Arturo Magidin May 10 '11 at 4:26
The action of $G$ on the left cosets of a maximal subgroup is primitive, and the subgroup is a stabilizer, so you can always realize any maximal subgroup as a stabilizer of some action. Did you mean that, or did you mean getting maximal subgroups as stabilizers of a fixed given action? – Arturo Magidin May 10 '11 at 4:30
yes..under what action, we get maximal subgroups as stabilizers? (It looks to be useful to get information about automorphism group of $p$ groups.) – user8186 May 10 '11 at 4:32
If $H$ is a maximal subgroup of $G$, $H$ is a stabilizer of the coset $H$ in the primitive action of $G$ on the left cosets of $H$, the action given by left multiplication. So every maximal subgroup is a stabilizer of a primitive action. – Arturo Magidin May 10 '11 at 4:34

To get an action of $S_3$ in which $\langle(1,2,3)\rangle$ is the stabilizer of a point, take $X=\{1,-1\}$, and let $S_3$ act by parity: even permutations fix both $1$ and $-1$, odd permutations swap $1$ and $-1$. Then the stabilizer of a single point, $1$, is $A_3 = \langle (1,2,3)\rangle$.

In general, if $H$ is a subgroup of $G$, then $G$ acts on the left cosets of $H$ by left multiplication, $g\cdot xH = gxH$. The stabilizer of the coset $H$ is $H$ itself. If you have a partition of the cosets so that $G$ acts on the partition, let $K = \{ g\in G\mid gH\text{ is in the same block as }H\}$ be the collection of all elements that represent cosets in the same part of the partition as $H$. Then $K$ certainly contains $H$. I claim that $K$ is a subgroup of $G$: for if $x\in K$, then $xH$ is the same block of the partition as $H$, hence $x^{-1}(xH)$ is in the same block as $x^{-1}H$. But $x^{-1}(xH) = H$, so $x^{-1}H$ is also in the same block of the partition, hence $x^{-1}\in K$. That is, $K$ is closed under inverses. And if $x,y\in K$, $x^{-1}\in K$. Taking the cosets $x^{-1}H$ and $yH$, which are in the same block, and multiplying by $x$, we conclude that $xyH$ and $H$ are in the same block, so $xy\in K$. Thus, $K$ is a subgroup, $H\leq K \leq G$.

Conversely, if $K$ is a subgroup, $H\leq K\leq G$, then $K$ induces a partition of the cosets into blocks for the action of $G$: just take the cosets of $K$, and partition each into cosets of $H$.

So systems of blocks in the action of $G$ on the left cosets of $H$ correspond to subgroups $K$, $H\leq K\leq G$.

Hence, if $H$ is a maximal subgroup, then the action of $G$ on the cosets is primitive, since the only possible systems of blocks are the trivial and total blocks, corresponding to $H$ and to $G$.

In particular, every maximal subgroup of $G$ is the stabilizer of a point in a primitive action of $G$, namely, $H$ is the stabilizer of $H$ in the (left) action of $G$ on the left cosets of $H$ in $G$.

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This is almost exactly Tits' algorithm. – Thomas Connor May 17 '11 at 16:10

Here is an older version of Arturo's answer:

The symmetric group of degree n acts on polynomials in n variables. We can view Sym(3) as acting on the variables x,y,z.

Consider its action on the polynomial x+y. The orbit contains x+y, x+z, and y+z. The stabilizer of x+y is the maximal subgroup Sym({x,y})≅Sym(2).

Consider its action on the polynomial Δ = (x−y)(x−z)(y−z). The orbit has only two polynomials, Δ and −Δ. The stabilizer of Δ is the maximal subgroup Alt({x,y,z})≅Alt(3).

You can play similar tricks with any subgroup of Sym(n). This is one way to look at "resolvents".

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