Let \(G\) be a group of order \(105 = 3 \cdot 5 \cdot 7\).
Suppose that \(G\) does not have a normal Sylow 7-subgroup. Show in this case that \(G\) has a normal Sylow 3-subgroup and a normal Sylow 5-subgroup. Prove then that \(G\) is abelian, a contradiction.
Show that there is a non-abelian group \(G\) of order 105. Explain why your group \(G\) is solvable, but not nilpotent.
Solution
Part (a).
1. If the Sylow \(7\)-subgroup is not normal, then \(n_7=15\).
Sylow’s theorem gives \[ n_7\mid 15, \qquad n_7\equiv1\pmod7. \] Hence \(n_7\in\{1,15\}\), and the hypothesis excludes \(1\).
2. The Sylow \(3\)-subgroup is normal.
Sylow’s theorem gives \(n_3\mid35\) and \(n_3\equiv1\pmod3\), so \(n_3\in\{1,7\}\). If \(n_3=7\), then the seven Sylow \(3\)-subgroups contribute \(7(3-1)=14\) nonidentity elements. By 1, the fifteen Sylow \(7\)-subgroups contribute \(15(7-1)=90\) nonidentity elements. Distinct subgroups of prime order meet only in the identity, so these sets are disjoint. Together with the identity this would give \[ 14+90+1=105 \] elements, leaving no element of order \(5\), contradicting Cauchy’s theorem. Thus \(n_3=1\).
3. The Sylow \(5\)-subgroup is normal.
Sylow’s theorem gives \(n_5\mid21\) and \(n_5\equiv1\pmod5\), so \(n_5\in\{1,21\}\). If \(n_5=21\), then the Sylow \(5\)-subgroups contribute \(21(5-1)=84\) nonidentity elements, while the Sylow \(7\)-subgroups contribute \(90\) by 1, impossible in a group of order \(105\). Hence \(n_5=1\).
4. If \(P_3\) and \(P_5\) denote the unique Sylow \(3\)- and \(5\)-subgroups, then\[ N=P_3P_5\cong C_{15} \]is a normal subgroup of \(G\).
Both subgroups are normal, and their coprime orders imply \(P_3\cap P_5=1\). For \(x\in P_3\) and \(y\in P_5\), the commutator \([x,y]\) belongs to both normal subgroups, hence to their trivial intersection. Thus they commute elementwise and \[ N=P_3\times P_5\cong C_3\times C_5\cong C_{15}. \] The product of normal subgroups is normal.
5. The conjugation action of \(G\) on \(N\) is trivial.
Because \(N\trianglelefteq G\), conjugation gives a homomorphism \[ G\longrightarrow \operatorname{Aut}(N). \] Since \(N\) is abelian, \(N\) lies in the kernel, so the action factors through \[ G/N\cong C_7. \] But \[ |\operatorname{Aut}(C_{15})|=\varphi(15)=8. \] The image of a group of order \(7\) in a group of order \(8\) must be trivial. Therefore every element of \(G\) centralizes \(N\).
6. The group \(G\) is abelian, a contradiction.
4. Hence all generators of \(G\) commute, so \(G\) is abelian.An abelian group has every subgroup normal, contradicting the assumption that its Sylow \(7\)-subgroup is not normal.
Part (b).
7. There is a nonabelian group of order \(21\) of the form\[ H=C_7\rtimes C_3. \]
Since \[ \operatorname{Aut}(C_7)\cong C_6, \] there is an automorphism of order \(3\). Using the corresponding nontrivial homomorphism \(C_3\to\operatorname{Aut}(C_7)\) gives a semidirect product \(H=C_7\rtimes C_3\). The action is nontrivial, so \(H\) is nonabelian.
8. The group\[ G=H\times C_5 \]is a nonabelian group of order \(105\).
Its order is \(21\cdot5=105\), and it contains the nonabelian factor \(H\).
9. The group \(G\) is solvable.
The subgroup \(C_7\trianglelefteq H\) has cyclic quotient \(C_3\), so \(H\) is solvable. Taking the direct product with the abelian group \(C_5\) preserves solvability. Equivalently, \[ 1\trianglelefteq C_7\trianglelefteq H\trianglelefteq H\times C_5 \] has abelian successive quotients \(C_7\), \(C_3\), and \(C_5\).
10. The group \(G\) is not nilpotent.
In the nontrivial semidirect product \(H=C_7\rtimes C_3\), a Sylow \(3\)-subgroup is not normal; otherwise both Sylow subgroups of \(H\) would be normal and, having coprime orders, would commute, forcing \(H\) to be the direct product \(C_7\times C_3\) and hence abelian. Therefore the corresponding Sylow \(3\)-subgroup of \(G=H\times C_5\) is not normal. Every finite nilpotent group has all Sylow subgroups normal, so \(G\) is not nilpotent.