Show that the transitive subgroups of \(S_3\) are \(S_3, A_3\)
Hint
By orbit-stabilizer, \(3\) divides the order of a transitive subgroup of \(S_3\).
Solution
1. Let \(H \leq S_3\) act transitively on \(X = \ts{1,2,3}\).
2. \(3\) divides \(\abs H\).
2.1. Transitivity says the orbit of \(1\) is all of \(X\), so it has \(3\) elements.
2.2. Orbit-stabilizer gives \(\abs H = 3 \cdot \abs{H_1}\), where \(H_1\) is the stabilizer of \(1\) in \(H\).
3. \(\abs H \in \ts{3, 6}\).
\(\abs H\) divides \(\abs{S_3} = 6\) by Lagrange, and step 2 rules out \(1\) and \(2\).
4. If \(\abs H = 6\) then \(H = S_3\).
\(H \leq S_3\) and the two have the same finite order.
5. If \(\abs H = 3\) then \(H = A_3\).
5.1. \(H\) is a Sylow \(3\)-subgroup of \(S_3\).
5.2. The number \(n_3\) of Sylow \(3\)-subgroups satisfies \(n_3 \equiv 1 \pmod 3\) and \(n_3 \mid 2\), so \(n_3 = 1\).
5.3. \(A_3 = \gens{(1\,2\,3)}\) has order \(3\), so it is that one Sylow subgroup, and \(H = A_3\).
6. Both \(S_3\) and \(A_3\) are transitive.
\((1\,2\,3) \in A_3\) carries \(1\) to \(2\) to \(3\), so the orbit of \(1\) under \(A_3\), and a fortiori under \(S_3\), is all of \(X\).
7. Q.E.D.
Steps 3 through 5 show a transitive subgroup is \(S_3\) or \(A_3\), and step 6 shows both are.