If you are looking to build your own "Overleaf" document, here is the code for a high-quality solution set covering selected exercises (4.1, 4.2, and 4.3).
You can copy and paste this directly into an Overleaf project.
\documentclass[12pt, a4paper]article
\usepackage[utf8]inputenc
\usepackagegeometry
\usepackageamsmath, amssymb, amsthm
\usepackageenumitem
\geometrymargin=1in
% Theorem Styles
\newtheorempropositionProposition
\newtheoremproblemProblem
\titleSolutions to Dummit \& Foote: Chapter 4\\Group Actions
\authorCompiled Solutions
\date\today
\begindocument
\maketitle
\sectionSection 4.1: Group Actions and Permutation Representations
\beginproblem[Exercise 4.1.1]
Let $G$ be a group acting on a set $A$. Prove that the relation $\sim$ defined by $a \sim b$ if and only if $b = g \cdot a$ for some $g \in G$ is an equivalence relation.
\endproblem
\beginproof
To show $\sim$ is an equivalence relation, we must verify reflexivity, symmetry, and transitivity.
\beginenumerate[label=(\roman*)]
\item \textbfReflexivity: Let $a \in A$. Since $G$ acts on $A$, $1 \cdot a = a$ for the identity element $1 \in G$. Thus, $a \sim a$.
\item \textbfSymmetry: Suppose $a \sim b$. Then there exists $g \in G$ such that $b = g \cdot a$. Since $G$ is a group, $g^-1 \in G$. Then:
\[ g^-1 \cdot b = g^-1 \cdot (g \cdot a) = (g^-1g) \cdot a = 1 \cdot a = a. \]
Thus, $a = g^-1 \cdot b$, which implies $b \sim a$.
\item \textbfTransitivity: Suppose $a \sim b$ and $b \sim c$. Then there exist $g, h \in G$ such that $b = g \cdot a$ and $c = h \cdot b$. Substituting, we get:
\[ c = h \cdot (g \cdot a) = (hg) \cdot a. \]
Since $hg \in G$, we have $a \sim c$.
\endenumerate
\endproof
\beginproblem[Exercise 4.1.3]
Show that the stabilizer $G_a$ of a point $a$ is a subgroup of $G$.
\endproblem
\beginproof
Let $G_a = \g \in G \mid g \cdot a = a\$.
\beginenumerate[label=(\roman*)]
\item \textbfIdentity: Since $1 \cdot a = a$, $1 \in G_a$.
\item \textbfClosed under inverses: If $g \in G_a$, then $g \cdot a = a$. Applying $g^-1$ to both sides:
\[ g^-1 \cdot (g \cdot a) = g^-1 \cdot a \implies 1 \cdot a = g^-1 \cdot a \implies a = g^-1 \cdot a. \]
Thus, $g^-1 \in G_a$.
\item \textbfClosed under products: If $g, h \in G_a$, then:
\[ (gh) \cdot a = g \cdot (h \cdot a) = g \cdot a = a. \]
Thus, $gh \in G_a$.
\endenumerate
Therefore, $G_a \le G$.
\endproof
\sectionSection 4.2: The Class Equation
\beginproblem[Exercise 4.2.1]
Let $G$ be a finite group of order $n$. Show that the size of the conjugacy class of an element $x \in G$ divides $n$.
\endproblem
\beginproof
The group $G$ acts on itself by conjugation. The orbit of an element $x$ under this action is its conjugacy class, denoted $\mathcalO_x$ or $\textCl(x)$. The stabilizer of $x$ is the centralizer $C_G(x) = \g \in G \mid gxg^-1 = x\$.
By the Orbit-Stabilizer Theorem:
\[ |\mathcalO_x| = [G : C_G(x)]. \]
The index $[G : C_G(x)]$ divides $|G| = n$ by Lagrange's Theorem. Therefore, the size of the conjugacy class divides $n$.
\endproof
\sectionSection 4.3: Group Actions on Sets
\beginproblem[Exercise 4.3.5]
Show that if $G$ is a group of order $p^2$ ($p$ prime), then $G$ is abelian.
\endproblem
\beginproof
The center of $G$, denoted $Z(G)$, is non-trivial for any $p$-group. Thus $|Z(G)|$ is either $p$ or $p^2$.
\beginenumerate
\item Suppose $|Z(G)| = p^2$. Then $Z(G) = G$, so $G$ is abelian.
\item Suppose $|Z(G)| = p$. Then the order of the quotient $G/Z(G)$ is $p$. Groups of prime order are cyclic. Let $G/Z(G) = \langle xZ(G) \rangle$.
Let $g, h \in G$. Then $gZ(G) = x^iZ(G)$ and $hZ(G) = x^jZ(G)$ for some $i,j$. This implies $g = x^i z_1$ and $h = x^j z_2$ for $z_1, z_2 \in Z(G)$.
Since elements in $Z(G)$ commute with everyone:
\[ gh = (x^i z_1)(x^j z_2) = x^i+j z_1 z_2. \]
\[ hg = (x^j z_2)(x^i z_1) = x^j+i z_2 z_1. \]
Since $x^i+j = x^j+i$ and $z_1 z_2 = z_2 z_1$, we have $gh = hg$. Thus $G$ is abelian.
\endenumerate
In either case, $G$ is abelian.
\endproof
\enddocument
Many professors post their own solution sets. Search for "Math 250A Dummit Foote solutions" – these often cover Chapter 4 in depth.
To create a professional solution manual, begin with this minimal Overleaf template:
\documentclass[12pt]article \usepackageamsmath, amssymb, amsthm \usepackageenumitem \usepackagetikz-cd \usepackagehyperref\newtheoremexerciseExercise[section] \theoremstyledefinition \newtheoremsolutionSolution dummit+and+foote+solutions+chapter+4+overleaf+full
\titleDummit & Foote Chapter 4 Solutions: Group Actions \authorYour Name \date\today
\begindocument
\maketitle \tableofcontents
\sectionGroup Actions and Permutation Representations If you are looking to build your own
\beginexercise[4.1.1] Let $G$ be a group and let $X$ be a set. Define a group action. \endexercise
\beginsolution A group action is a map $G \times X \to X$, denoted $(g,x) \mapsto g \cdot x$, satisfying... \endsolution
% Continue for each exercise \enddocument
Organize solutions by subsection (4.1, 4.2, ..., 4.5 for Sylow Theorems). Use \label and \ref to reference previous exercises—common in Chapter 4, where later exercises build on orbit decompositions.
Before diving into solutions, let's understand the landscape. Chapter 4 is structured as follows:
The problems in Chapter 4 are infamous. They include:
A student who masters Chapter 4’s exercises has internalized the very essence of group theory. But the official Dummit and Foote solutions are not publicly endorsed by the authors (to preserve pedagogical integrity). Instead, the community has built meticulous, crowd-sourced solutions. Many professors post their own solution sets