- Classify all abelian groups of order 36 up to isomorphism.
For the rest of the problem, assume that \(G\) is a non-abelian group of order 36. You may assume that the only subgroup of order 12 in \(S_4\) is \(A_4\) and that \(A_4\) has no subgroup of order 6.
Prove that if the Sylow 2-subgroup of \(G\) is normal, then \(G\) has a normal subgroup \(N\) such that \(G/N \cong A_4\).
Show that if \(G\) has a normal subgroup \(N\) such that \(G/N \cong A_4\) and a subgroup \(H \cong A_4\), then \(G\) must be the direct product of \(N\) and \(H\).
Show that the dihedral group of order 36 is a non-abelian group of order 36 whose Sylow 2-subgroup is not normal.
Solution
Goal: Classify abelian groups of order 36, prove that non-abelian groups with normal Sylow 2-subgroups admit an \(A_4\) quotient, prove direct product decomposition, and exhibit \(D_{18}\) as a non-abelian counterexample to Sylow 2 normality.
1. Part (a): Classification of abelian groups of order 36.
1.1. The prime factorization of 36 is \(36 = 2^2 \cdot 3^2\).
1.2. By the Fundamental Theorem of Finitely Generated Abelian Groups, any finite abelian group is isomorphic to a direct product of cyclic groups of prime power orders.
1.3. Partitions of the prime exponents:- For \(p = 2\): the partitions of 2 are \(2\) and \(1 + 1\), corresponding to \(\mathbb{Z}/4\mathbb{Z}\) and \(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}\).- For \(p = 3\): the partitions of 2 are \(2\) and \(1 + 1\), corresponding to \(\mathbb{Z}/9\mathbb{Z}\) and \(\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}\).
1.4. There are \(2 \times 2 = 4\) non-isomorphic abelian groups of order 36:1. \(\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z} \cong \mathbb{Z}/36\mathbb{Z}\).2. \(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/18\mathbb{Z}\).3. \(\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/12\mathbb{Z}\).4. \(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}\).
2. Part (b): If \(P \in \operatorname{Syl}_2(G)\) is normal, then \(G/N \cong A_4\) for some normal subgroup \(N \trianglelefteq G\).
2.1. Let \(P \in \operatorname{Syl}_2(G)\) with \(|P| = 4\) and \(P \trianglelefteq G\).
2.2. Number of Sylow 3-subgroups: By Sylow’s Theorems, \(n_3 \equiv 1 \pmod 3\) and \(n_3 \mid 4\), so \(n_3 \in \{1, 4\}\).
2.3. If \(n_3 = 1\), let \(Q \in \operatorname{Syl}_3(G)\) (\(|Q| = 9\)). Then \(Q \trianglelefteq G\).- Since \(P \trianglelefteq G\) and \(Q \trianglelefteq G\) with \(\gcd(|P|, |Q|) = \gcd(4, 9) = 1\), \(P \cap Q = \{e\}\) and \(G \cong P \times Q\).- Any group of order \(p^2\) is abelian, so both \(P\) and \(Q\) are abelian.- The direct product of abelian groups is abelian, so \(G\) is abelian, contradicting the hypothesis that \(G\) is non-abelian.
2.4. Thus \(n_3 = 4\).
2.5. Conjugation action on \(\operatorname{Syl}_3(G)\):- \(G\) acts transitively on the set \(X = \operatorname{Syl}_3(G)\) of size 4 by conjugation.- This defines a permutation homomorphism \(\rho: G \to \operatorname{Sym}(X) \cong S_4\).
2.6. Kernel and image of \(\rho\):- Let \(Q \in X\). The stabilizer is \(\operatorname{Stab}_G(Q) = N_G(Q)\), which has index \([G : N_G(Q)] = n_3 = 4\), so \(|N_G(Q)| = 36/4 = 9\), which forces \(N_G(Q) = Q\).- The kernel \(N = \ker(\rho) = \bigcap_{g \in G} g Q g^{-1} \subseteq Q\) is a 3-group, so \(|N| \in \{1, 3, 9\}\).- The image \(\operatorname{Im}(\rho) \cong G/N\) has order \(\frac{36}{|N|}\) and is a subgroup of \(S_4\), so \(\frac{36}{|N|}\) divides \(|S_4| = 24\).- Thus \(\frac{36}{|N|}\) divides 24, which requires \(3 \mid |N|\).- If \(|N| = 9\), then \(N = Q \trianglelefteq G\), which would mean \(n_3 = 1\), a contradiction.- Hence \(|N| = 3\).
2.7. Identification of the quotient:- \(|\operatorname{Im}(\rho)| = \frac{36}{3} = 12\).- Since \(\operatorname{Im}(\rho)\) is a subgroup of order 12 in \(S_4\), by the given assumption \(\operatorname{Im}(\rho) = A_4\).- Therefore \(G/N \cong A_4\), where \(N = \ker(\rho) \trianglelefteq G\) is of order 3.
3. Part (c): Direct product decomposition \(G \cong N \times H\).
3.1. We are given \(N \trianglelefteq G\) with \(G/N \cong A_4\) (\(|N| = 3\)) and \(H \le G\) with \(H \cong A_4\) (\(|H| = 12\)).
3.2. Trivial intersection \(N \cap H = \{e\}\):- \(N \cap H\) is a normal subgroup of \(H \cong A_4\) whose order divides \(|N| = 3\).- The normal subgroups of \(A_4\) are \(\{e\}\), \(V_4\) (order 4), and \(A_4\) (order 12); \(A_4\) has no normal subgroup of order 3.- Thus \(N \cap H = \{e\}\).
3.3. Full group product:- \(|N H| = \frac{|N| |H|}{|N \cap H|} = \frac{3 \cdot 12}{1} = 36 = |G|\).- Therefore \(G = N H\).
3.4. Commutativity of elements:- Conjugation defines a homomorphism \(\psi: H \to \operatorname{Aut}(N)\).- Since \(|N| = 3\), \(\operatorname{Aut}(N) \cong (\mathbb{Z}/3\mathbb{Z})^\times \cong \mathbb{Z}/2\mathbb{Z}\).- The commutator subgroup \([H, H] = [A_4, A_4] = V_4\), so the abelianization of \(H\) is \(H^{\text{ab}} \cong A_4/V_4 \cong \mathbb{Z}/3\mathbb{Z}\).- Since \(\gcd(3, 2) = 1\), there are no non-trivial homomorphisms from \(A_4\) to \(\mathbb{Z}/2\mathbb{Z}\), so \(\psi\) is trivial.- Thus \(h n h^{-1} = n\), i.e., \(n h = h n\) for all \(n \in N\) and \(h \in H\).
3.5. Hence \(G\) is the internal direct product \(G \cong N \times H \cong \mathbb{Z}/3\mathbb{Z} \times A_4\).
4. Part (d): \(D_{18}\) is non-abelian of order 36 with non-normal Sylow 2-subgroups.
4.1. The dihedral group of order 36 has presentation:\[D_{18} = \langle r, s \mid r^{18} = 1, s^2 = 1, s r s = r^{-1} \rangle.\]
4.2. \(|D_{18}| = 36\). Since \(s r = r^{17} s \ne r s\), \(D_{18}\) is non-abelian.
4.3. A Sylow 2-subgroup has order 4.
4.4. Consider the subgroup \(P = \{e, r^9, s, s r^9\} \le D_{18}\), which has order 4.
4.5. Conjugate \(P\) by \(r\):\[r s r^{-1} = s r^{-2} = s r^{16}.\]
4.6. Since \(s r^{16} \notin P\) (\(16 \not\equiv 0, 9 \pmod{18}\)), \(r P r^{-1} \ne P\).
4.7. Thus the Sylow 2-subgroup \(P\) is not normal in \(D_{18}\).
5. Conclusion:
Abelian groups of order 36 are classified by integer partitions, normal Sylow 2-subgroups yield an $A_4$ quotient via conjugation on $\operatorname{Syl}_3(G)$, subgroups isomorphic to $A_4$ decompose $G$ as a direct product $\mathbb{Z}/3\mathbb{Z} \times A_4$, and $D_{18}$ has non-normal Sylow 2-subgroups.