I am reading the book "Permutation Groups" by Dixon and Mortimer in which they discuss blocks and primitivity of group actions.

An important theorem which I just read its proof states:

Let $G$ act transitively on a nonempty set $\Omega$ and let $\alpha \in \Omega$. Denote $\mathcal B$ the set of all blocks containing $\alpha$ and $\mathcal S$ all subgroups of $G$ that contain $G_\alpha$ (the stabilizer of $\alpha$). Then there exists a bijection $\Psi : \mathcal B \rightarrow \mathcal S$ given by $\Psi (\Delta)=G_ {(\Delta)}$ (the point stabilizer of the block $\Delta$). The inverse map $\Phi=\Psi^{-1} $ is given by $\Phi(H)=H.\alpha$ (the orbit of $\alpha$ under the restricted action of $H$). Moreover, this map is order preserving.

The exercise I'm having some trouble with is to find all blocks of $G=\langle (123456),(35)(26)\rangle =\langle x,y\rangle$ that include 1 (that is $\mathcal B$) and identify them with the corresponding subgroups (that is the elements of $\mathcal S$)

The first part is simple and I think there is no escape from going through all cases- of course we have the trivial blocks ${1}$ and $\{1,2,3,4,5,6\}$. For 2 elemented blocks observe that $\{1,\beta\}$ can be seperated by $x$ or by $x^2$ whenever $\beta=2,3,5,6$ though if $\beta=4$ so both x,y preserve the system $\{\{1,4\}\{2,5\}\{3,6\}\}$ and therefore any word of x,y will preserve this system and hence $\{1,4\}$ is a block. For 3 elemented blocks observe that $x$ seperates every $\{1,\beta,\gamma\}$ unless $\beta=3,\gamma=5$ and indeed in that case both $x$ and $y$ preserve the system $\{\{1,3,5\},\{2,4,6\}\}$ and hence $\{1,3,5\}$ is a block.

Now I need to identify the corresponding subgroups which are $G_{(1,4)}$ and $G_{(1,3,5)}$. Any suggestions how to do such? I've tried several words of $x$ and $y$ to find some pattern but no success in actually finding the exact subgroup and proving that indeed that's the correct subgroup.

Thanks a lot.


$\newcommand{\Span}{\langle #1 \rangle}$You've done an excellent job.

Perhaps it's worth noting that since $y = (3 5) (2 6)$ inverts by conjugation $x = (123456)$ (that is, $y^{-1} x y = x^{-1}$), the group has order $12$, and its elements can be written uniquely in the form $x^{j} y^{i}$, for $0 \le i < 2$ and $0 \le j < 6$. (This is used in (sub) below.)

Clearly the stabilizer of $1$ is then $\langle y \rangle$ (it fixes $1$, and it has the right index $6$), and the subgroups containing it are in one-to-one correspondence with the subgroups of $\langle x\rangle$, which is a cyclic group of order $6$, that is, the subgroups containing $\langle y \rangle$ are $$ \langle y \rangle, \langle x^{2}, y \rangle, \langle x^{3}, y \rangle, G. $$

Of course you could have computed these subgroups first, and then the blocks as the orbits of $1$ under these subgroups. In particular you would have obtained proper, non-trivial blocks as $$\tag{sub} \langle x^{2}, y \rangle \cdot 1 = \langle x^{2} \rangle \cdot 1 = \{1, x^{2}, x^{4} \} \cdot 1 = \{1, x^{2} \cdot 1, x^{4} \cdot 1\} = \{ 1, 3, 5 \}, \\ \langle x^{3}, y \rangle \cdot 1 = \langle x^{3} \rangle \cdot 1 = \{1, x^{3} \cdot 1 \} = \{ 1, 4 \}. $$

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