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    The transitive subgroups of $S_3$ are $S_3$ and $A_3$

    P-IFQ4C

    Area
    Algebra
    Topics
    Group Actions, Permutations, Subgroups
    Status
    draft

    Show that the transitive subgroups of the symmetric group \(S_3\) (acting naturally on \(\{1, 2, 3\}\)) are precisely \(S_3\) and \(A_3\).

    Solution

    Let \(H\le S_3\) act transitively on \(\{1,2,3\}\). By orbit–stabilizer, \[ |H|=3\,|H_1|, \] so \(3\mid |H|\). Since \(|H|\mid6\), we have \(|H|=3\) or \(6\).

    If \(|H|=6\), then \(H=S_3\). If \(|H|=3\), then \(H\) is generated by a \(3\)-cycle. The two \(3\)-cycles generate the same subgroup \[ A_3=\{1,(123),(132)\}, \] which is transitive. Hence the transitive subgroups are exactly \(A_3\) and \(S_3\).

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