Draw the subgroup lattice for \(S_3\).
Solution
The subgroups of \(S_3\) are \[ \{e\}, \qquad \langle(12)\rangle, \qquad \langle(13)\rangle, \qquad \langle(23)\rangle, \qquad A_3=\langle(123)\rangle, \qquad S_3. \]
1. These are all the subgroups of \(S_3\).
By Lagrange’s theorem, a subgroup has order dividing \(6\), hence order \(1,2,3\), or \(6\). There is one subgroup of order \(1\) and one of order \(6\). A subgroup of order \(2\) is generated by an element of order \(2\); the three transpositions give exactly the three displayed subgroups. A subgroup of order \(3\) is generated by a \(3\)-cycle, and the two \(3\)-cycles generate the same subgroup \(A_3\). Thus the list is exhaustive.
2. The subgroup lattice is\[ \begin{array}{c} S_3\\[6pt] /\quad|\quad|\quad\\\\[-6pt] A_3\quad \langle(12)\rangle\quad \langle(13)\rangle\quad \langle(23)\rangle\\[6pt] \backslash\quad|\quad|\quad/\\\\[-6pt] \{e\}. \end{array} \]
Each order-\(2\) or order-\(3\) subgroup is maximal because its index is prime. Distinct subgroups of prime order intersect trivially, and none of the order-\(2\) subgroups lies in \(A_3\). Thus the only strict inclusions are from \(\{e\}\) to the four proper nontrivial subgroups and from each of those to \(S_3\).
3. Among the proper nontrivial subgroups, only \(A_3\) is normal.
The subgroup \(A_3\) has index \(2\), hence is normal. The three transposition subgroups are conjugate to one another and therefore none is normal.