Let $G$ act by conjugation on its Sylow 7-subgroup $P_7$. Since $|P_7| = 7$, $|\operatornameAut(P_7)| = 6$. This action induces a homomorphism $G \to \operatornameAut(P_7)$. The kernel of this map is $C_G(P_7)$, and we have $G / C_G(P_7) \le \operatornameAut(P_7)$. Hence $|G / C_G(P_7)|$ divides $6$. But $G / C_G(P_7)$ is a group of order $105 / |C_G(P_7)|$. This forces $|C_G(P_7)| = 105$ or $35$ or $21$ or $15$ or $7$ or $5$ or $3$ or $1$, but careful analysis shows a contradiction. The standard approach is to consider that $G$ acting by conjugation on $P_7$ gives a homomorphism $\varphi: G \to S_n_7$. The rest of the proof involves counting arguments that force $n_5 = 1$. The details are left as an exercise for the reader (or you can look up the full proof).
Navigating Abstract Algebra: A Guide to Dummit and Foote Chapter 4 Solutions on Overleaf dummit+and+foote+solutions+chapter+4+overleaf+full
: Some solutions are extremely rigorous, while others might skip "obvious" algebraic manipulations, which can be frustrating for someone seeing the material for the first time. Technical Quality Mathematical Notation : Uses standard packages like , ensuring that symbols like is congruent to (isomorphism) and \trianglelefteq (normal subgroup) are rendered correctly. Let $G$ act by conjugation on its Sylow 7-subgroup $P_7$