Let \(G\le S_n\) act transitively on the \(n\) letters.
- Prove that \(n\) divides \(|G|\).
- Suppose \(n=4\). For which integers \(k\ge1\) can such a transitive subgroup have order \(4k\)?
Solution
1. One has\[ n\mid |G|. \]
Fix one of the \(n\) letters, say \(x\). Since the action is transitive, the orbit of \(x\) has cardinality \(n\). By the orbit-stabilizer theorem, \[ |G| = |G\cdot x|\,|G_x| = n|G_x|. \] Hence \(n\) divides \(|G|\).
2. If \(n=4\) and \(|G|=4k\), then\[ k\in\{1,2,3,6\}. \]
Because \(G\le S_4\), Lagrange’s theorem gives \[ |G|\mid |S_4|=24. \] Thus \(4k\mid24\), so \(k\mid6\). Since \(k\ge1\), this gives \(k\in\{1,2,3,6\}\).
3. Every value in step 2 occurs for a transitive subgroup of \(S_4\).
3.1. The value \(k=1\) occurs.
The cyclic subgroup \[ \langle(1234)\rangle \] has order \(4\) and is transitive because the orbit of \(1\) under the powers of \((1234)\) is all of \(\{1,2,3,4\}\).
3.2. The value \(k=2\) occurs.
Let \[ r=(1234), \qquad s=(13). \] Then \(r^4=s^2=1\) and \(srs=r^{-1}\). Hence every element of \(\langle r,s\rangle\) can be written as \(r^i\) or \(sr^i\) with \(0\le i<4\), so this subgroup has at most eight elements. These eight elements are distinct: the four powers of \(r\) are distinct, and an equality \(sr^i=r^j\) would imply \(s=r^{j-i}\), whereas \((13)\) is not a power of the four-cycle \(r\). Thus \[ |\langle r,s\rangle|=8. \] It is transitive because it contains \(r\).
3.3. The value \(k=3\) occurs.
The alternating group \(A_4\) has order \(12\). Its natural action on the four letters is transitive: for each \(j\ne1\), a three-cycle containing \(1\) and \(j\) sends \(1\) to \(j\).
3.4. The value \(k=6\) occurs.
The full symmetric group \(S_4\) has order \(24\) and acts transitively on its four letters.
3.5. Q.E.D.
Steps <2>1–<2>4 exhibit transitive subgroups of orders \(4\), \(8\), \(12\), and \(24\), respectively.
4. Therefore the possible integers are exactly\[ \boxed{k\in\{1,2,3,6\}}. \]
Step 2 shows that no other value can occur, and step 3 shows that every listed value does occur.
5. Q.E.D.
Step 1 proves part (1), and step 4 proves part (2).