Constructing a particular transitive subgroup

Question

Using Cayley’s theorem, I want to construct a transitive subgroup of $S_6$ which is isomorphic to $S_3$. I've been struggling with group theory rather a lot, and I find this particular problem to be very difficult.
Cayley's Theorem states that: "Every group $G$ is isomorphic to a subgroup of $S_G$". What exactly does this imply for the problem that I want to solve? I know what a transitive group is, but I don't know how Cayley's Theorem comes into this problem.
On a related note, does one develop an intuition for group theory with enough practice? I've had enormous issues with the general abstractions so far. Thanks in advance.

Answer 1

Hint: There are six elements of $S_3$, and the elements in $S_3$ canonically act on these by ...

Answer 2

It is not a bad idea to make this abstract algebra less abstract.
The subgroup of $S_6$ generated by $\sigma=(1\,2\,3)(4\,5\,6)$ and $\tau=(1\,4)(2\,5)(3\,6)$ is the symmetry group of a triangular prism, and it is pretty obviously transitive and isomorphic to $S_3$.

Answer 3

(Multiplication is right-to-left.)

By the orbit-stabilizer theorem, all the pointwise stabilizers are trivial, so every non-identity permutation of the sought subgroup, say $K$, moves all the elements of $\{1,\dots,6\}$. Therefore, the elements of order $3$ and $2$ which generate $K$ must have cycle type $(3,3)$ and $(2,2,2)$, respectively. Choose as the former $\sigma=(123)(456)$, whence $\sigma^2=(132)(465)$. Now, let's look for a suitable element of order $2$ and cycle type $(2,2,2)$, say $\tau=(1i)(jk)(lm)$, considered that a presentation for $K\cong S_3\cong D_3$ is: $$K=\langle\sigma,\tau\mid(\tau\sigma)^2=()\rangle\tag1$$ We get $i\ne2,3$, because otherwise $\tau\sigma$ or $\sigma\tau$, respectively, fixes $1$. So, let's take $\tau=(14)(2k)(lm)$. Clearly $k\ne 3$, because otherwise $\tau\sigma$ fixes $2$. Therefore, $\tau=(14)(25)(36)$ or $\tau=(14)(26)(35)$. But the latter only fulfils $(1)$, and hence a suitable $K$ is: $$K=\langle(123)(456),(14)(26)(35)\rangle$$

Answer 4

Using Cayley’s theorem, I want to construct a transitive subgroup of $S_6$ which is isomorphic to $S_3$.

So thoroughly formulated, I see no shortcuts to the complete routine.

Cayley's embedding $\varphi\colon S_3\hookrightarrow S_{S_3}$ is defined by: \begin{alignat}{1} &()&&\mapsto ()\circ\_ \\ &(12)&&\mapsto (12)\circ\_ \\ &(13)&&\mapsto (13)\circ\_ \\ &(23)&&\mapsto (23)\circ\_ \\ &(123)&&\mapsto (123)\circ\_ \\ &(132)&&\mapsto (132)\circ\_ \\ \end{alignat} To embed the transitive subgroup $(S_3\cong)\varphi(S_3)\le S_{S_3}$ into $S_6$, you've got to choose any amongst the $6!$ bijections $S_3\to \{1,\dots,6\}$, say for example $f$ defined by: \begin{alignat}{1} &()&&\mapsto 1 \\ &(12)&&\mapsto 2 \\ &(13)&&\mapsto 3 \\ &(23)&&\mapsto 4 \\ &(123)&&\mapsto 5 \\ &(132)&&\mapsto 6 \\ \tag1 \end{alignat} Then, the isomorphism $\psi_f\colon S_{S_3}\to S_6$, defined by $\sigma\mapsto f\sigma f^{-1}$, does the job. For example: \begin{alignat}{1} (\psi_f(()\circ\_))[1] &= (f(()\circ\_)) f^{-1})[1] \\ &= f((()\circ\_)(f^{-1}[1])) \\ &= f(()\circ ()) \\ &= f(()) \\ &= 1 \end{alignat} \begin{alignat}{1} (\psi_f(()\circ\_))[2] &= (f(()\circ\_)) f^{-1})[2] \\ &= f((()\circ\_)(f^{-1}[2])) \\ &= f(()\circ (12)) \\ &= f((12)) \\ &= 2 \end{alignat} \begin{alignat}{1} (\psi_f(()\circ\_))[3] &= (f(()\circ\_)) f^{-1})[3] \\ &= f((()\circ\_)(f^{-1}[3])) \\ &= f(()\circ (13)) \\ &= f((13)) \\ &= 3 \end{alignat} etc.. So, finally: $$\psi_f(()_{S_3}\circ\_)=()_{S_6}$$ Likewise, you will find: \begin{alignat}{1} &\psi_f((12)\circ\_) &&=(12)(36)(45) \\ &\psi_f((13)\circ\_) &&=(13)(25)(46) \\ &\psi_f((23)\circ\_) &&=(14)(26)(35) \\ &\psi_f((123)\circ\_) &&=(156)(234) \\ &\psi_f((132)\circ\_) &&=(165)(243) \\ \end{alignat} whence: $$S_3\stackrel{\psi_f\varphi}{\cong}\{(),(12)(36)(45),(13)(25)(46),(14)(26)(35),(156)(234),(165)(243)\}\le S_6$$ Other choices of the bijection $f$ in $(1)$ will lead to different transitive subgroups of $S_6$ isomorphic to $S_3$.

(Btw, for $f$ defined by $()\mapsto 1$, $(123)\mapsto 2$, $(132)\mapsto 3$, $(12)\mapsto 4$, $(13)\mapsto 5$, $(23)\mapsto 6$, you will get the subgroup as in other answer of mine.)