Chapter 4 Overleaf - Dummit And Foote Solutions

\sectionGroup Actions and Permutation Representations

\beginabstract This document presents rigorous solutions to selected exercises from Chapter 4 of Dummit and Foote's \textitAbstract Algebra, Third Edition. The focus is on group actions, orbit-stabilizer theorem, $p$-groups, and applications to Sylow theory. Each solution emphasizes clear reasoning and formal justification. \endabstract Dummit And Foote Solutions Chapter 4 Overleaf

\beginsolution Let $n_p$ and $n_q$ be the numbers of Sylow $p$- and $q$-subgroups. By Sylow, $n_p \equiv 1 \pmodp$ and $n_p \mid q$. Since $p \neq q$, $n_p = 1$ or $n_p = q$. Similarly, $n_q \equiv 1 \pmodq$ and $n_q \mid p^2$, so $n_q = 1, p, p^2$. If $n_p = 1$, the Sylow $p$-subgroup is normal and we are done. If $n_q = 1$, done. Assume $n_p = q$ and $n_q \neq 1$. Then $n_q = p$ or $p^2$. But $n_q \equiv 1 \pmodq$ forces $p \equiv 1 \pmodq$ or $p^2 \equiv 1 \pmodq$. These conditions contradict $p,q$ distinct and the counting of elements (each Sylow $q$-subgroup contributes $q-1$ non-identity elements, etc.). A standard counting argument shows $n_p = 1$ must hold. \endsolution \endabstract \beginsolution Let $n_p$ and $n_q$ be the

\sectionConclusion and Further Directions Similarly, $n_q \equiv 1 \pmodq$ and $n_q \mid

\beginexercise[Section 4.5, Exercise 3] Let $G$ be a finite group, $p$ a prime, and let $P$ be a Sylow $p$-subgroup of $G$. Prove that $N_G(N_G(P)) = N_G(P)$. \endexercise

\beginsolution Apply the class equation: [ |G| = |Z(G)| + \sum_i [G : C_G(g_i)], ] where the sum runs over non-central conjugacy classes. Each $[G : C_G(g_i)] > 1$ is a power of $p$ (since $C_G(g_i)$ is a subgroup). Thus $p$ divides each term in the sum. Also $p \mid |G|$. Hence $p \mid |Z(G)|$. Therefore $|Z(G)| \geq p$, so $Z(G)$ is nontrivial. \endsolution

\beginsolution Consider the action of $G$ on itself by left multiplication. This gives a homomorphism $\varphi: G \to S_2n$. However, a more refined approach uses Cayley's theorem and parity.