\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
\beginexercise[Section 4.2, Exercise 8] Let $G$ be a $p$-group acting on a finite set $A$. Prove that [ |A| \equiv |\Fix(A)| \pmodp, ] where $\Fix(A) = a \in A : g \cdot a = a \text for all g \in G$. \endexercise
\beginsolution Let $H = N_G(P)$. By definition, $P \triangleleft H$ (since $P$ is normal in its normalizer). Hence $P$ is the unique Sylow $p$-subgroup of $H$. Now let $g \in N_G(H)$. Then $gPg^-1 \subseteq gHg^-1 = H$, so $gPg^-1$ is also a Sylow $p$-subgroup of $H$. By uniqueness, $gPg^-1 = P$. Thus $g \in N_G(P) = H$. Therefore $N_G(H) \subseteq H$, and the reverse inclusion is trivial. So $N_G(H) = H$. \endsolution
\documentclass[12pt, leqno]article \usepackage[utf8]inputenc \usepackageamsmath, amssymb, amsthm, amscd \usepackage[margin=1in]geometry \usepackageenumitem \usepackagetitlesec \usepackagexcolor % -------------------------------------------------------------- % Custom Commands for Dummit & Foote Notation % -------------------------------------------------------------- \newcommand\Z\mathbbZ \newcommand\R\mathbbR \newcommand\C\mathbbC \newcommand\Q\mathbbQ \newcommand\F\mathbbF \newcommand\Stab\textStab \newcommand\Fix\textFix \newcommand\Orb\textOrb \newcommand\sgn\textsgn \newcommand\Aut\textAut \newcommand\Inn\textInn \newcommand\soc\textSoc \newcommand\Ker\textKer \newcommand\Image\textIm Dummit And Foote Solutions Chapter 4 Overleaf
\beginsolution Let $n_3$ denote the number of Sylow $3$-subgroups. By Sylow's theorems, $n_3 \equiv 1 \pmod3$ and $n_3 \mid 4$. The divisors of $4$ are $1,2,4$. Which are $\equiv 1 \pmod3$? $1 \equiv 1 \pmod3$, $4 \equiv 1 \pmod3$, but $2 \equiv 2 \pmod3$. Hence $n_3 = 1$ or $n_3 = 4$. No other possibilities. \endsolution
\beginexercise[Section 4.1, Exercise 3] Let $G$ be a group and let $H \leq G$. Prove that the action of $G$ on the set of left cosets $G/H$ by left multiplication is transitive. Determine $\Stab_G(1H)$. \endexercise
\beginexercise[Section 4.2, Exercise 2] Let $G$ act on a finite set $A$. Prove that if $G$ acts transitively on $A$, then $|A|$ divides $|G|$. \endexercise \beginsolution Let $n_p$ and $n_q$ be the numbers
\beginexercise[Section 4.4, Exercise 12] Let $G$ be a group of order $p^2q$ with $p$ and $q$ distinct primes. Prove that $G$ has a normal Sylow subgroup. \endexercise
\sectionGroup Actions and Permutation Representations
\sectionConclusion and Further Directions Similarly, $n_q \equiv 1 \pmodq$ and $n_q \mid
\sectionThe Class Equation and Consequences
\sectionGroup Actions on Sylow Subgroups