Subgroup of $S_6$ which is isomorphic to $S_3$ I have to construct a transitive subgroup of $S_6$ which is isomorphic to $S_3$.
I was thinking of using Cayley's theorem, but I have no clue how to construct such a transitive subgroup.
I know the order of $S_3$ is 6 and the order of $S_6$ is 720
Any help would be grateful. Thanks in advance.
 A: Label $S_3=\{\sigma_1,\cdots,\sigma_6 \}$. Each $\sigma_i$ determines a permutation of $S_3$ by $l_i : \sigma_j \mapsto \sigma_i \sigma_j$. This gives a one-to-one homomorphism $S_3 \hookrightarrow S_6$.
$S_6$ acts on $S_3$ by $\tau \cdot \sigma_j=\sigma_{\tau(j)}$. Each $\sigma_i \in S_3$ defines a permutation $\tau_i$ of $\{1,\cdots,6\}$.
Moreover, for any $\sigma_j$ and $\sigma_k$ in $S_3$, taking $\sigma=\sigma_k\sigma_j^{-1}$ one has $\sigma \sigma_j=\sigma_k$, that is the action of $S_3$ (as a subgroup of $S_6$) is transitive on $\{ \sigma_1,\cdots,\sigma_6 \}$.
A: Consider the map from $S_3$ to $S_6$ which sends:
$$Id \to Id$$
$$(1,2)\to (1,4)(2,5)(3,6)$$
$$(1,3)\to (1,5)(2,6)(3,4)$$
$$(2,3)\to (1,6)(2,4)(3,5)$$
$$(1,2,3) \to (1,3,2)(4,5,6)$$
$$(1,3,2)\to (1,2,3)(4,6,5)$$
This is a homomorphism given by the Cayley theorem. The corresponding action is transitive.
Note:  when we multiply permutations $ab$ the permutation $a$ acts first.
A: We want $S_3$ to act transitively on 6 points. Now $|S_3| = 6$ and a group acts transitively on its own elements.
Let us make this explicit, label each of the elements of $S_3$ with a number from 1 to 6:

*

*$\color{green}1 = (\color{blue}{})$

*$\color{green}2 = (\color{blue}{1\,2})$

*$\color{green}3 = (\color{blue}{1\,3})$

*$\color{green}4 = (\color{blue}{2\,3})$

*$\color{green}5 = (\color{blue}{1\,2\,3})$

*$\color{green}6 = (\color{blue}{1\,3\,2})$
We understand the action of $S_3$ on the blue objects and we can use it to write out the action of the green objects. For example $(1\,2)$ acts like so:
gap> (1,2)*();
(1,2)      { which is green 2 }
gap> (1,2)*(1,2);
()         { which is green 1 }
gap> (1,2)*(1,3);
(1,2,3)    { which is green 5 }
gap> (1,2)*(2,3);
(1,3,2)    { which is green 6 }
gap> (1,2)*(1,2,3);
(1,3)      { which is green 3 }
gap> (1,2)*(1,3,2);
(2,3)      { which is green 4 }

Therefore $\sigma (1\,2) = (\color{green}{1\,2})(\color{green}{3\,5})(\color{green}{4\,6})$ and you can compute the $S_6$ permutation representation of each element of $S_3$ in a similar way. Giving you a subgroup with a transitive action on 6 points.
A: A subgroup of order $6$ of $S_6$, say $\Sigma$, which acts transitively on the set $X:=\{1,\dots,6\}$, must have all the stabilizers trivial (Orbit-Stabilizer Theorem): $ \forall i\in X, \operatorname{Stab}(i)=\{Id_X\}$, or equivalently: $ \forall i\in X, \sigma\in \Sigma\mid\sigma(i)=i\Rightarrow \sigma=Id_X$. Therefore, we need to look at the elements of $S_6$ which do not fix any element of $X$, namely:

*

*$6$-cycles;

*products of a $2$-cycle with a disjoint $4$-cycle;

*products of $2$ disjoint $3$-cycles;

*products of $3$ disjoint $2$-cycles.

Now, narrowing the survey to $\Sigma$'s isomorphic to $S_3$, options 1 and 2 are ruled out because $S_3$ is made of (further than the identity) $3$ elements of order $2$ and $2$ elements of order $3$. Therefore, any closed $\Sigma\subseteq S_6$ made of the identity, three products of $3$ disjoint $2$-cycles, and two products of $2$ disjoint $3$-cycles, is a transitive subgroup isomorphic to $S_3$.
