Define a solvable group. Give an example of a solvable nonabelian group.
Show \(A_4\) is solvable. Do the Sylow theorems tell you anything about whether this index 3 subgroup of \(A_4\) is normal?
Solution
1. Definition of a solvable group:
1.1. A group \(G\) is called solvable if it possesses a finite subnormal series:\[ \{e\} = G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_k = G \]such that each consecutive quotient group \(G_{i+1}/G_i\) is abelian for all \(i \in \{0, \dots, k-1\}\).Equivalently, the derived series \(G^{(0)} = G\), \(G^{(n+1)} = [G^{(n)}, G^{(n)}]\) satisfies \(G^{(k)} = \{e\}\) for some \(k \ge 0\).
2. Example of a solvable nonabelian group:
2.1. The symmetric group on 3 elements \(S_3\) has order \(|S_3| = 6\).It is nonabelian (e.g. \((12)(23) \neq (23)(12)\)) and has the subnormal series:\[ \{e\} \triangleleft A_3 \triangleleft S_3, \]where \(A_3 \cong \mathbb{Z}/3\mathbb{Z}\) and \(S_3/A_3 \cong \mathbb{Z}/2\mathbb{Z}\) are both abelian.
3. Proof that \(A_4\) is solvable:
3.1. The alternating group \(A_4\) has order \(|A_4| = 12\).Consider the Klein four-subgroup:\[ V_4 = \{ e, (12)(34), (13)(24), (14)(23) \} \le A_4. \]
3.2. Conjugation in \(S_4\) preserves the cycle type of permutations, so \(V_4 \triangleleft S_4\), which implies \(V_4 \triangleleft A_4\).
3.3. Form the subnormal series:\[ \{e\} \triangleleft V_4 \triangleleft A_4. \]The quotient factors are:- \(V_4 / \{e\} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}\) (abelian),- \(A_4 / V_4\) has order \(\frac{12}{4} = 3\), so \(A_4 / V_4 \cong \mathbb{Z}/3\mathbb{Z}\) (abelian).Thus \(A_4\) is solvable.
4. Sylow theorems and normality of the index-3 subgroup:
4.1. The subgroup \(V_4\) has order \(|V_4| = 4 = 2^2\).Since \(|A_4| = 12 = 2^2 \cdot 3\), \(V_4\) is a Sylow 2-subgroup of \(A_4\).
4.2. By the Sylow Theorems, the number \(n_2\) of Sylow 2-subgroups of \(A_4\) satisfies:\[ n_2 \mid 3 \quad \text{and} \quad n_2 \equiv 1 \pmod 2 \implies n_2 \in \{1, 3\}. \]
4.3. \(A_4\) contains exactly 8 elements of order 3 (the eight 3-cycles) and exactly 3 elements of order 2 (the three double transpositions in \(V_4\)).Since any Sylow 2-subgroup of order 4 contains 3 elements of order 2, and there are only 3 elements of order 2 in the entire group \(A_4\), there cannot be more than one Sylow 2-subgroup.Thus \(n_2 = 1\).
4.4. By Sylow theory, a Sylow \(p\)-subgroup is normal if and only if \(n_p = 1\).Therefore, \(V_4\) is the unique Sylow 2-subgroup of \(A_4\), and is consequently a normal subgroup of \(A_4\).
5. Conclusion:\(A_4\) is solvable, and the Sylow theorems together with element counts establish that its unique Sylow 2-subgroup (the index-3 Klein four-group) is normal. Q.E.D.